Advanced Mathmatics
First Edition
1
Selected Lectures on Inequalities
Part
In real-world scenarios and everyday life, a myriad of equal and unequal relationships exist, including
more/less, big/small, long/short, tall/short, far/near, fast/slow, up/down, light/heavy, not more than/not
less than. These issues are manifested in the numerical relationships between quantities, namely equality and
inequality. Equations express equality, while inequalities express inequality.
Sec 1.1
1.1.1
Fundamentals of inequality
Relationship between two real numbers
Firstly, we review the fundamental facts regarding the correlation between the values of two real numbers.
Since the points on the numerical axis correspond to the real numbers in a one-to-one manner, the relationship
between the positions of the points on the axis can be utilized to specify the relationship between the values
of the real numbers as follows: let a, b be two real numbers, and their corresponding points on the numerical
axis are A, B. Then
A
a
B
b
B
b
x
When A is to the left of B, a < b
A
a
x
When A is to the right of B, a > b
The following are the basic facts about the comparison of the values of the real numbers a, b:
If a − b is positive, then a > b; if a − b is equal to 0, then a = b; if a − b is negative, then a < b. And vice
versa.
It follows from the aforementioned basic facts that we can compare the values of two real numbers
by comparing their difference with respect to the magnitude of 0. This method is fundamental in proving
inequalities, known as the comparison of differences.
Comparison of differences A ≥ B holds if A − B ≥ 0.
Example 1.1.1.
Suppose x ∈ R, compare the values of (x + 2)(x + 3) and (x + 1)(x + 4).
Solution.
(x + 2)(x + 3) − (x + 1)(x + 4) = (x2 + 5x + 6) − (x2 + 5x + 4) = 2 > 0,
hence (x + 2)(x + 3)>(x + 1)(x + 4).
1
2
ADVANCED MATHMATICS
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1.1. FUNDAMENTALS OF INEQUALITY
1.1.2
Basic Properties of Inequalities
We begin by recalling some basic properties of inequalities, for any a, b, c, d ∈ R, the following hold:
1. a ≥ b ⇔ b ≤ a;
2. If a ≥ b, c ≥ d, then a + c ≥ b + d;
3. If a ≥ b ≥ 0, c ≥ d ≥ 0, then ac ≥ bd;
4. a ≥ b ⇔ a + c ≥ b + c;
a
a+c
b
b+c
x
b+c
b
c>0
a+c
a
x
c<0
5. If c > 0, then a ≥ b ⇔ ac ≥ bc.
6. If c < 0, then a ≥ b ⇔ ac ≤ bc.
We can derive some simple propositions from the above properties.
Proposition 1.1.1.
If a + b ≥ c, then a ≥ c − b, where a, b, c ∈ R.
This proposition is, in fact, the theoretical underpinning of what is usually referred to as the “shift term”.
Proof. By property 4 above, it holds that a + b ≥ c ⇒ a + b − b ≥ c − b ⇒ a ≥ c − b.
Proposition 1.1.2.
If a ≥ b ≥ 0, then a2 ≥ b2 .
Proof. By property 3 above, let c = a, d = b, then a · a ≥ b · b, that is, a2 ≥ b2 .
Remark:
1. In the conclusion and proof above, we can analyze further: from a ≥ b ≥ 0, we get a2 ≥
b2 ≥ 0. Of course, using the property of inequalities 3 again, if we take a2 as c and b2 as d, we get
a3 ≥ b3 ≥ 0, and so on, by mathematical induction we get
Proposition 1.1.3.
If a ≥ b ≥ 0, then an ≥ bn ≥ 0, where n ∈ N+ ;
2. If a, b ≥ 0, by the monotonically increasing properties of function f (x) = x n and an ≥ bn ≥ 0, we get
1
a ≥ b ≥ 0.
1.1.3
Common problems about inequalities
1.1.3.1 Proving inequalities
Proving an inequality is essentially a comparison of two numbers.
1.1. FUNDAMENTALS OF INEQUALITY
3
In higher mathematics, there are numerous problems related to proving inequalities, including integral
methods and ideas for proving inequalities remain basically consistent. For instance, the fundamental method
of comparing two real numbers, as previously mentioned, involves comparing differences. Additionally, another
significant method of comparison involves comparing quotients.
A
Comparison of quotients If B > 0, then A ≥ B if B
≥ 1.
However, these two comparison methods are typically the initial step in establishing an inequality and
require further processing using other techniques. One classic and significant approach is deflation. In some
cases, we can directly demonstrate that the inequality A ≥ B is more challenging. To address this, we can
attempt to identify an intermediate number C. If both A ≤ C and C ≤ B hold simultaneously, it naturally
follows that A ≤ B. By employing deflation, we refer to the process of zooming from A to C and then from
C to B, or vice versa, zooming from B to C and then to A. The art of proving inequalities often lies in the
extent of deflation.
Here we give a classic example of the deflation method.
Example 1.1.2.
If a, b ∈ R, then
|a + b|
|a|
|b|
≤
+
.
1 + |a + b|
1 + |a| 1 + |b|
Proof.
x
Step 1: Let f (x) = 1+x
, then the readers can prove that f (x) is increasing on [0, +∞);
Step 2: By triangle inequality, it holds that |a + b| ≤ |a| + |b|;
Step 3: By the above two steps,
f (|a + b|) =
|a + b|
|a| + |b|
≤ f (|a| + |b|) =
1 + |a + b|
1 + |a| + |b|
=
|b|
|a|
+
1 + |a| + |b| 1 + |a| + |b|
≤
|a|
|b|
+
.
1 + |a| 1 + |b|
The reader is presumably conversant with the aforementioned methodologies delineated in prior studies.
Furthermore, the methods of proving inequalities are analytical method, the method of undetermined coefficients and the standardization method and so on. For more ideas and methods of proving inequalities, please
reference [1].
1.1.3.2 Solving inequalities
Solving an inequality essentially translates to discerning a solution set to an inequality with variables.
Throughout our research, we might confront higher-order, absolute value, or even fractional inequalities, each
necessitating distinct treatment methods. However, our focus here is solely on resolving quadratic inequalities.
The standardized process for solving a quadratic inequality parallels that of solving a quadratic equation,
concisely delineated in a five-step sequence.
1. Remove the denominator.
1
SELECTED LECTURES ON INEQUALITIES
inequalities, which are classic and important. Despite the various complexities in higher mathematics, the
4
1.1. FUNDAMENTALS OF INEQUALITY
2. Remove parentheses.
ADVANCED MATHMATICS
1
3. Shift terms;
4. Merge terms;
5. Combine terms
6. Divide both sides of an inequality by the coefficient of the unknown to obtain the solution set of the
inequality. (Note that, however, the sign of the inequality changes direction when both sides of the
inequality are multiplied or divided by a negative number.)
The main theoretical basis for these steps is the basic properties of inequalities mentioned at the beginning
of the subsection 1.1.2.
Example 1.1.3.
Solve the inequality 1+x
≥ 2x−5
, and represent its solution set on the numerical axis.
2
−3
Solution.
Remove the denominator, multiplied by −6 , then
−3(1 + x) ≤ 2(2x − 5),
Remove parentheses, then
−3 − 3x ≤ 4x − 10,
Shift terms, then
−3x − 4x ≤ −10 + 3.
Merge terms, then
−7x ≤ −7,
Divided by −7 , then x ≥ 1. The solution set of this inequality is represented on the numerical axis as
follows.
−2
1.1.4
−1
0
1
2
3
4
x
More inequalities
As advancements in mathematical exploration persist, an increasing number of inequalities are being validated. The book [2] encapsulates these inequalities from various mathematical branches, including fundamental inequalities, number theory and combinatorial inequalities, algebraic inequalities, geometric inequalities,
elementary transcendental functional inequalities, polynomial inequalities, convex functions, and variational
inequalities.
The ensuing section will primarily concentrate on three significant inequalities: Bernoulli’s, the mean
value, and Cauchy’s inequalities.
Sec 1.2
1.2.1
Three important inequalities
Bernoulli’s inequality
Theorem 1.2.1.
If x > −1, x ̸= 0, n ∈ N+ and n ≥ 2, then
(1 + x)n > 1 + nx.
In discerning the implications of Bernoulli’s inequality, it appears to suggest that exponential functions
exert a form of “control” over primary functions, and by extension, polynomial functions. Viewed from the lens
of limits, without prerequisite knowledge on the subject but merely a rudimentary understanding, consider a
polynomial f (x) = a0 + a1 x + · · · + an xn with real coefficients (where ai ∈ R, i = 0, 1, · · · , n and an ̸= 0), all
of which maintain the subsequent values
f (x)
= 0.
x→+∞ ex
lim
Such a framework implies that the polynomial function becomes relatively insignificant compared to the
exponential function as x increases to a high degree.
An examination of the function plot underscores this relationship: the exponential function demonstrates
swift, significant growth, distinctly surpassing the polynomial function for relatively “large” x values. While
formalizing this observation with standard high school knowledge may not be viable, it becomes feasible to
articulate this finding more precisely following the application of limit theory.
Subsequently, a proof of Bernoulli’s inequality will be presented.
Proof.
Method 1: (Binomial Theorem) By the Binomial Theorem
(1 + x)n = 1 + nx +
n(n − 1) 2
x + · · · + xn > 1 + nx,
2!
where x ̸= 0, n ∈ N+ and n ≥ 2.
Method 2: (Mathematical induction)
Step 1: When n = 2 ,
(1 + x)2 = 1 + 2x + x2
> 1 + 2x
(x ̸= 0, so inequality holds strictly.)
Step 2: Assuming that the conclusion holds when n = k, then when n = k + 1,
(1 + x)k+1 = (1 + x)k (1 + x)
> (1 + kx)(1 + x) (induction hypothesis and x > −1)
= 1 + (k + 1)x + kx2
> 1 + (k + 1)x
i.e. conclusion also holds when n = k + 1.
(Because x ̸= 0)
5
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SELECTED LECTURES ON INEQUALITIES
1.2. THREE IMPORTANT INEQUALITIES
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ADVANCED MATHMATICS
1
1.2. THREE IMPORTANT INEQUALITIES
Step 3: By Step 1 and Step 2 we get that when x > −1, x ̸= 0, n ∈ N+ and n ≥ 2, it holds that
(1 + x)n > 1 + nx.
1.2.2
Mean value inequality
Theorem 1.2.2.
If a1 , a2 , · · · , an ≥ 0, n ∈ N+ and n ≥ 2, then
√
a1 + a2 + · · · + an
≥ n a1 a2 · · · · an ,
n
and “=” holds if and only if a1 = a2 = · · · = an .
The mean value inequality, an integral component of high school curricula and predominantly known
in its n = 2 manifestation, is presumably familiar to many readers. It is often cited in some high school
textbooks as the basic inequality or the mean value inequality.
Theorem 1.2.3.
If a, b ≥ 0, then
a+b √
≥ ab,
2
and “=” holds if and only if a = b.
Another common form is the square, or so-called real (not necessarily nonnegative) form.
Theorem 1.2.4.
If a, b ∈ R, then
a 2 + b2
≥ ab,
2
and “=” holds if and only if a = b.
There are many ways to prove the mean value inequality, for example, twelve are given in the book [3].
We will give two proofs below, both of which are essentially mathematical inductions.
While mathematical induction may be a topic familiar from junior high and high school, it remains a
crucial methodology within the domain of advanced mathematics. Consequently, it is imperative for the
reader to attain mastery in its application. Furthermore, the less frequently encountered technique known
as “forward-backward mathematical induction”rarely surfaced in high school curricula, thus, this serves as an
opportune moment to delve into its utilization.
Proof.
Method 1: (Mathematical induction)
Step 1: When n = 2 ,
a1 + a2 √
− a1 a2 =
2
√
√ 2
a1 − a2
≥ 0,
2
hence
a1 + a2 √
≥ a1 a2 , and
2
“=”holds ⇐⇒
√
a1 −
√
a2 = 0 ⇐⇒ a1 = a2 .
Step 2: Suppose that conclusion is true when n = k , then when n = k + 1 , WLOG, assume that
ak+1 = max {a1 , a2 , · · · , ak+1 } .
then
a1 + a2 + · · · + ak+1
k+1
k+1
a1 + a2 + · · · + ak
kak+1 − (a1 + a2 + · · · + ak )
=
+
k
k(k + 1)
=(A + B)k+1 ≥ Ak+1 + (k + 1)Ak B
k+1
(Binomial theorem)
=Ak [A + (k + 1)B] ≥ a1 a2 · · · ak · ak+1 ,
(induction hypothesis)
hence,
a1 + a2 + · · · + ak+1
√
≥ k+1 a1 a2 · · · ak+1 ,
k+1
and by the condition for the equality sign in the binomial theorem and in the induction hypothesis
“=” holds ⇐⇒ a1 = a2 = · · · = ak+1 .
Step 3: By Step 1 and Step 2, we get that if a1 , a2 , · · · , an > 0, n ∈ N+ and n ≥ 2, then
√
a1 + a2 + · · · + an
≥ n a1 a2 · · · · an ,
n
and “=” holds if and only if a1 = a2 = · · · = an .
Method 2: (Forward-backward mathematical induction)
Remark: Most individuals lack familiarity with the “forward-backward mathematical induction,”
a concept often left untouched in high school curricula and referred to as backward induction or
“backfill” in [3]. The principle methodology involves initially validating the proposition for a subset,
{nm }, of natural numbers, then proceeding to confirm the truth of the case {n}\ {nm }.
Step 1: When n = 2 ,
a1 + a2 √
− a1 a2 =
2
so
√
√ 2
a1 − a2
≥ 0,
2
a1 + a2 √
≥ a1 a2 , and
2
“=” holds ⇐⇒
√
a1 −
√
a2 = 0 ⇐⇒ a1 = a2 .
Step 2: Forward induction. Assuming that the conclusion holds when n = 2k , then when n = 2k+1 , it holds
7
1
SELECTED LECTURES ON INEQUALITIES
1.2. THREE IMPORTANT INEQUALITIES
8
1.2. THREE IMPORTANT INEQUALITIES
that
ADVANCED MATHMATICS
1
a1 + a2 + · · · + a2k+1
2k+1
1 a 1 + a 2 + · · · + a 2k
a2k +1 + a2k +2 + · · · + a2k+1
=
+
2
2k
2k
1 2√
√
k
a1 a2 · · · a2k + 2k a2k +1 a2k +2 · · · a2k+1
(induction hypothesis)
2
q
√
√
2k
≥
a1 a2 · · · a2k · 2k a2k +1 a2k +2 · · · a2k+1 (By Step 1)
≥
√
= 2k+1 a1 a2 · · · a2k+1
and by Step1 and induction hypothesis,
“=”holds if and only if a1 = · · · = a2k = a2k +1 = · · · = a2k+1 .
i.e. the conclusion is true when n = 2k+1 .
Step 3: Backward induction. Assuming that the conclusion holds when n = m, then when n = m − 1, it
holds that
a1 + a2 + · · · + am−1
m−1
=
m
1
·
(a1 + a2 + · · · + am−1 )
m m−1
+···+am−1
a1 + a2 + · · · + am−1 + a1 +a2m−1
=
m
r
a1 + a2 + · · · + am−1
≥ m a1 a2 · · · am−1 ·
m−1
(induction hypothesis)
Perform the m power on both sides, then
hence
a1 + a2 + · · · + am−1
m−1
m
≥ a1 a2 · · · am−1 ·
a1 + a2 + · · · + am−1
m−1
a1 + a2 + · · · + am−1
,
m−1
m−1
≥ a1 a2 · · · am−1
so,
a1 + a2 + · · · + am−1
√
≥ m−1 a1 a2 · · · am−1 ,
m−1
‘‘ =′′ holds if and only if a1 = a2 = · · · = am−1 .
i.e. the conclusion is true when n = m − 1.
Step 4: By Step 1, Step 2 and Step 3, we get that if a1 , a2 , · · · , an > 0, n ∈ N+ and n ≥ 2, then
√
a1 + a2 + · · · + an
≥ n a1 a2 · · · an
n
holds and “=” holds if and only if a1 = a2 = · · · = an .
An application of the mean value inequality in proving inequalities is given below:.
1.2. THREE IMPORTANT INEQUALITIES
Example 1.2.4.
a
b
c
+
+ ≥ 2.
a b+c c
Analysis: Applying the mean value inequality directly for n = 3, it becomes evident that the product of the
three left-hand terms is non-constant. The core issue lies in the non-cancellation between the denominator
of the second term, b + c, and the numerator of the third term, b. Furthermore, the denominator presents
complexities in handling. To circumvent this, one strategy is to approximate b as b+c in the numerator, thereby
facilitating the application of the mean value inequality and consequently substantiating the proposition.
Proof. By the mean value inequality,
c
a
b
c
a
b+c
+
+ = +
+
−1
a b+c c
a b+c
c
r
a
b+c
c
≥33 ·
·
−1
a b+c
c
= 3 − 1 = 2.
Cauchy’s inequality
If you have encountered math competitions or pursued elective high school courses, you are likely familiar
with Cauchy’s inequality. While not a staple in most high school curricula, comprehending and mastering its
fundamental applications can greatly aid in proving various inequalities.
Theorem 1.2.5.
If a1 , a2 , · · · , an ∈ R, b1 , b2 , · · · , bn ∈ R, n ∈ N+ and n ≥ 2, then
(
n
X
ai bi )2 = (a1 b1 + a2 b2 + · · · + an bn )
i=1
≤ a21 + a22 + · · · + a2n
2
b21 + b22 + · · · + b2n ,
and “=” holds if and only if there exist constants that are not all zero λ, µ, such that
λ · (a1 , a2 , · · · , an ) = µ · (b1 , b2 , · · · , bn )
We first give a proof of Cauchy’s inequality, then some comments on this inequality, and finally conclude
with an example of using Cauchy’s inequality to prove an inequality.
Like the proof of the mean value inequality, Cauchy’s inequality also has a lot of ways to prove it, for
example, [3] book gives fifteen ways to prove it, we give two of them here.
Proof.
Method 1: (Discriminant method)
Pn
Case 1: If k=1 a2k = 0, the inequality holds apparently.
SELECTED LECTURES ON INEQUALITIES
1
([3, P19, Example 5]) Suppose a, b, c > 0, prove that
1.2.3
9
10
1.2. THREE IMPORTANT INEQUALITIES
Case 2: If
ADVANCED MATHMATICS
1
Pn
2
k=1 ak > 0, by
2
(ak t − bk ) ≥ 0,
we have
n
X
!
t −
a2k
2
2
k=1
n
X
!
a k bk
t+
k=1
hence
∆=
∀1 ≤ k ≤ n,
−2

= 4
n
X
n
X
−4
k=1
n
X
∀t ∈ R
!
!
n
X
a2k
k=1
!2
−
a k bk
k=1
n
X
!2
a k bk
≤
k=1
n
X
n
X
a2k
k=1
!
b2k  ≤ 0.
k=1
!
a2k
b2k
k=1
!
k=1
i.e.
n
X
b2k ≥ 0,
k=1
!2
a k bk
n
X
∀t ∈ R
n
X
!
b2k
,
k=1
and the equality holds if and only if there exists t0 ∈ R, s.t.
b k = t0 a k ,
∀1 ≤ k ≤ n.
Method 2: (Normalization using homogeneity)
Pn
Pn
Case 1: If k=1 a2k = 0 or k=1 b2k = 0, the conclusion holds apparently.
Pn
Pn
Case 2: If k=1 a2k = An > 0 and k=1 b2k = Bn > 0, then let
bk
ak
xk = √ , y k = √ ,
An
Bn
∀1 ≤ k ≤ n
then
n
X
x2k =
k=1
n
X
yk2 = 1,
k=1
hence the original proposition is equivalent to
|x1 y1 + x2 y2 + · · · + xn yn | ≤ 1.
And
|x1 y1 + x2 y2 + · · · xn yn | ≤ |x1 y1 | + |x2 y2 | + · · · + |xn yn |
x21 + y12 x22 + y22
x2 + yn2
+
+ ··· + n
2
2
2
!
n
n
1 X 2 X 2
yk = 1
=
xk +
2 k=1
k=1
≤
and the equality holds if and only if x1 y1 , x2 y2 , · · · , xn yn have the same sign, and
|xk | = |yk | ,
∀1 ≤ k ≤ n,
i.e.
a
b
√k = √k ,
An
Bn
ak
bk
∀1 ≤ k ≤ n or √
= −√ ,
An
Bn
∀1 ≤ k ≤ n.
1.2. THREE IMPORTANT INEQUALITIES
1. We usually rewrite
(
n
X
ai bi )2 = (a1 b1 + a2 b2 + · · · + an bn )
1
2
i=1
≤ a21 + a22 + · · · + a2n
b21 + b22 + · · · + b2n ,
as
n
X
1
1
ai bi ≤ a21 + a22 + · · · + a2n 2 b21 + b22 + · · · + b2n 2
i=1
n
X
=
! 12
|ai |2
i=1
n
X
! 12
|bi |2
i=1
2. As evident from the previously elaborated proof, we can establish a more potent inequality:
n
X
|ai bi | ≤
i=1
n
X
! 21
|ai |
2
n
X
! 12
|bi |
2
.
(1.1)
i=1
i=1
Why is this inequality stronger? Because by the absolute value inequality we know that
n
X
a i bi ≤
n
X
i=1
|ai bi |,
(1.2)
i=1
if inequality (1.1) holds, then the Cauchy inequality also holds immediately by inequality (1.2).
3. (a) Indeed, implementing the second method requires merely a minor modification to render the conclusion perfectly congruous with the instance of complex numbers. (Prompt: Consider that direct
comparison is unfeasible for complex numbers, thereby disallowing straightforward application of
the mean value inequality; however, the modulus of complex numbers is a real number).
(b) In the case of complex numbers (and thus also in the case of real numbers), the Cauchy inequality
is a strict characterization of the difference between the two terms, which is called the Lagrange’s
identity (readers wishing for further information may conduct their own exploration).
4. We see that in the Cauchy inequality
(a) the power of ai and the power of the sum are inversely related, i.e. 2 · 21 = 1;
Pn
Pn
(b) the powers of the two i=1 a2i and i=1 b2i satisfy 21 + 12 = 1,
if we generalize these two observations to the general case, we get the Hölder inequality:
Theorem 1.2.6.
If a1 , a2 , · · · , an ∈ R, b1 , b2 , · · · , bn ∈ R, n ∈ N+ and n ≥ 2, then for any p > 1, q > 1 satisfying
1
+ 1q = 1, the following inequality holds:
p
n
X
i=1
a i bi ≤
n
X
i=1
! p1
|ai |
p
n
X
! q1
|bi |
q
.
i=1
Interested readers can look up the proof of the Hölder inequality themselves.
SELECTED LECTURES ON INEQUALITIES
Remark:
11
12
1.2. THREE IMPORTANT INEQUALITIES
5. The Cauchy inequality is valid for any n > 1, signifying its applicability regardless of the magnitude of
ADVANCED MATHMATICS
1
n, be it tens of thousands or millions. Despite the extensive range of n, it remains a definite positive
integer, implying the existence of infinitely many integers exceeding it. One might ponder whether this
n could emulate a rocket with boundless thrust, perpetually ascending until it reaches the conceptual
’end of the universe’ - infinity ∞. Affirmatively, this notion aligns with the concept of limit and series
in advanced mathematics, expressed as
∞
X
an .
n=1
At this juncture, a comprehensive understanding of series is not crucial; rather, it suffices to recognize
that the aforementioned concept of ’infinite increase’ is pertinent to the Cauchy inequality. This relevance
becomes clearer as one delves deeper into series theory.
Theorem 1.2.7.
If ai , bi ∈ R, i = 1, 2, · · · , then the following inequality holds:
∞
X
a i bi ≤
i=1
∞
X
! 12
|ai |2
∞
X
! 12
|bi |2
.
i=1
i=1
Similarly, there is a corresponding series form of the Hölder inequality mentioned in the above comment:
Theorem 1.2.8.
If ai , bi ∈ R, i = 1, 2, · · · , then for any p > 1, q > 1 satisfying p1 + 1q = 1, the following inequality
holds:
∞
X
a i bi ≤
i=1
∞
X
! p1
|ai |
p
∞
X
! q1
|bi |
q
.
i=1
i=1
6. The finite or series constructs discussed above are inherently “discrete”. Naturally, there exist also
“continuous” variants: the integral forms of the Cauchy inequality and the Hölder inequality. These
forms will be delved into during subsequent explorations of advanced mathematics. As a result, a
detailed exposition is abstained from in this context. Readers possessing keen interest may refer to
pertinent advanced mathematical textbooks for further enrichment.
Finally, we give a classic application of Cauchy’s inequality in proving inequalities:
Example 1.2.5.
Given a > 0, b > 0, a + 2b = 1. Prove that:
1 2
+ ≥ 9.
a b
Proof. By Cauchy’s inequality,
1 2
1
4
+ = ( + )(a + 2b) =
a b
a 2b
≥
1
√
a
2
+
1 √
√ a+
a
2
√
2b
2 ! 2 √
√
2b
2b
√ 2 √ 2
a + 2b
2
= (1 + 2)2 = 9.
1.2. THREE IMPORTANT INEQUALITIES
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Remark: We can get the following inequality by transforming the above inequality:
Can you infer a general conclusion? (Hint: the numerator needs to be written in square form)
1
SELECTED LECTURES ON INEQUALITIES
1 2
12 22
(1 + 2)2
+ =
+
≥
.
a b
a
2b
a + 2b
14
ADVANCED MATHMATICS
1
1.2. THREE IMPORTANT INEQUALITIES
References
[1] 苏勇, 熊斌. 不等式的解题方法与技巧[M]. 华东师范大学出版社, 2012.
[2] 匡继昌. 常用不等式 (第四版)[M]. 山东科学技术出版社, 2010.
[3] 李胜宏, 边红平. 平均值不等式与柯西不等式[M]. 华东师范大学出版社, 2012.
15
What we know is not much,
but what we do not know is immense.
-Pierre-Simon Laplace-