Algebra Worksheet 3 – Inequalities
1. For all real values of the variables involved, and using the Trivial Inequality:
a. Prove that 𝑥 2 + 2𝑥𝑦 + 𝑦 2 ≥ 0.
b. Hence or otherwise, prove that (𝑎 + 𝑏)2 + 2𝑎2 + (𝑎 − 𝑏)2 ≥ 2𝑏 2
2. Show that (𝑎2 𝑏 + 𝑏 2 𝑐 + 𝑐 2 𝑎)(𝑎𝑏 2 + 𝑏𝑐 2 + 𝑐𝑎2 ) ≥ 9𝑎2 𝑏 2 𝑐 2 for all real 𝑎, 𝑏, 𝑐 using the
AM-GM inequality.
𝑥
𝑦
𝑧
3. Show that 𝑦 + 𝑧 + 𝑥 ≥ 3 for all real 𝑥, 𝑦, 𝑧.
4. Given that 𝑥, 𝑦, 𝑧 are positive reals and that 𝑥𝑦𝑧 = 1, use both the AM-GM inequality and
Cauchy-Schwarz to prove that 𝑥 2 + 𝑦 2 + 𝑧 2 ≥ 𝑥 + 𝑦 + 𝑧
5. Given that 𝑎, 𝑏 and 𝑐 are positive real numbers, prove using the Cauchy-Schwarz inequality
that
𝑎2
𝑐
+
𝑏2
𝑎
+
𝑐2
𝑏
≥𝑎+𝑏+𝑐
6. [Source: UKMT Mentoring] Let 𝑛 be a positive integer. Let 𝑑(𝑛) denote the number of
divisors of 𝑛, and let 𝜎(𝑛) be the sum of these divisors (in each case including 𝑛 itself). Prove
that:
𝜎(𝑛)
≥ √𝑛
𝑑(𝑛)
(Hint: we have a sum here, so we can probably use the AM-GM inequality)
7. [Source: BMO1] Show that for all real 𝑎, 𝑏, 𝑐:
(𝑎2 + 𝑏 2 )2 ≥ (𝑎 + 𝑏 + 𝑐)(𝑎 + 𝑏 − 𝑐)(𝑏 + 𝑐 − 𝑎)(𝑐 + 𝑎 − 𝑏)
(Hint: Using the Trivial Inequality will do here. The hard part is working out the factorisation
after you’ve expanded!)
8. [Optional] [Source: Irish Olympiad 99] Let 𝑎, 𝑏, 𝑐, 𝑑 be positive real numbers which add to 1.
Prove that:
𝑎2
𝑏2
𝑐2
𝑑2
1
+
+
+
≥
𝑎+𝑏 𝑏+𝑐 𝑐+𝑑 𝑑+𝑎 2
(Hint: the squared 𝑎2 , 𝑏 2 , … terms on the LHS gives a clue as to what we should use on the
RHS of a Cauchy-Schwarz inequality)
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Algebra Worksheet 3 - ANSWERS
1.
For all real values of the variables involved, and using the Trivial Inequality:
a. Prove that 𝒙𝟐 + 𝟐𝒙𝒚 + 𝒚𝟐 ≥ 𝟎.
This factorises to (𝑥 + 𝑦)2 ≥ 0, which is true by the Trivial Inequality.
b. Hence or otherwise, prove that (𝒂 + 𝒃)𝟐 + 𝟐𝒂𝟐 + (𝒂 − 𝒃)𝟐 ≥ 𝟐𝒃𝟐
Rearranging: (𝑎 + 𝑏)2 + 2(𝑎2 − 𝑏)2 + (𝑎 − 𝑏)2 ≥ 0
Making the substitution 𝑥 = 𝑎 + 𝑏 and 𝑦 + 𝑎 − 𝑏, we obtain the inequality in (a).
2.
Show that (𝒂𝟐 𝒃 + 𝒃𝟐 𝒄 + 𝒄𝟐 𝒂)(𝒂𝒃𝟐 + 𝒃𝒄𝟐 + 𝒄𝒂𝟐 ) ≥ 𝟗𝒂𝟐 𝒃𝟐 𝒄𝟐 for all real 𝒂, 𝒃, 𝒄 using the AM-GM
inequality.
3
Using AM-GM on each factor: 𝑎2 𝑏 + 𝑏 2 𝑐 + 𝑐 2 𝑎 ≥ 3 √𝑎3 𝑏 3 𝑐 3 = 3𝑎𝑏𝑐. Similarly 𝑎𝑏 2 + 𝑏𝑐 2 + 𝑐𝑎2 ≥
3
3 √𝑎3 𝑏 3 𝑐 3 = 3𝑎𝑏𝑐. Multiplying these together yields the above inequality.
3.
𝒙
𝒚
𝒛
𝒚
𝒛
𝒙
Show that + + ≥ 𝟑 for all real 𝒙, 𝒚, 𝒛.
𝑥
𝑦
𝑧
3 𝑥𝑦𝑧
By the AM-GM inequality, + + ≥ 3 × √
=3
𝑦
𝑧
𝑥
𝑦𝑧𝑥
4.
Given that 𝒙, 𝒚, 𝒛 are positive reals and that 𝒙𝒚𝒛 = 𝟏, use both the AM-GM inequality and CauchySchwarz inequality to prove that 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 ≥ 𝒙 + 𝒚 + 𝒛.
By the AM-GM inequality, 𝑥 + 𝑦 + 𝑧 ≥ 3 3√𝑥𝑦𝑧 = 3.
By the Cauchy-Schwarz inequality, (1 + 1 + 1)(𝑥 2 + 𝑦 2 + 𝑧 2 ) ≥ (𝑥 + 𝑦 + 𝑧)2 .
So 3(𝑥 2 + 𝑦 2 + 𝑧 2 ) ≥ (𝑥 + 𝑦 + 𝑧)(𝑥 + 𝑦 + 𝑧) ≥ 3(𝑥 + 𝑦 + 𝑧).
Thus 𝑥 2 + 𝑦 2 + 𝑧 2 ≥ 𝑥 + 𝑦 + 𝑧.
5.
Given that 𝒂, 𝒃 and 𝒄 are positive real numbers, prove using the Cauchy-Schwarz inequality that
𝒂𝟐
𝒄
+
𝒃𝟐
𝒂
+
𝒄𝟐
𝒃
≥𝒂+𝒃+𝒄
By Cauchy-Schwarz:
𝑎2 𝑏 2 𝑐 2
+ + ) ≥ (𝑎 + 𝑏 + 𝑐)2
𝑐
𝑎
𝑏
But since (as specified by the question), 𝑎 + 𝑏 + 𝑐 > 0, we can divide both sides by 𝑎 + 𝑏 + 𝑐, giving
the desired inequality.
(𝑎 + 𝑏 + 𝑐) (
6.
Let 𝒏 be a positive integer. Let 𝒅(𝒏) denote the number of divisors of 𝒏, and let 𝝈(𝒏) be the sum of
these divisors (in each case including 𝒏 itself). Prove that:
𝝈(𝒏)
𝒅(𝒏)
≥ √𝒏
Suppose that the divisors of 𝑛 are 𝑛1 , 𝑛2 up to 𝑛𝑑(𝑛) .
Then the left hand side is
GM inequality,
𝑛1 +𝑛2 +⋯+𝑛𝑑(𝑛)
𝑛1 +𝑛2 +⋯+𝑛𝑑(𝑛)
𝑑(𝑛)
𝑑(𝑛)
≥
which is the Arithmetic Mean of the divisors. Thus by the AM-
𝑑(𝑛)
√𝑛1 𝑛2 … 𝑛𝑑(𝑛) . But we can organise the divisors into pairs such
𝑑(𝑛)
that the product of each pair is 𝑛, thus we find
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𝑑(𝑛)
√𝑛1 𝑛2 … 𝑛𝑑(𝑛) =
√𝑛
𝑑(𝑛)
2
= (𝑛
1
𝑑(𝑛) 𝑑(𝑛)
2
)
= √𝑛.
𝟐
7. Show that for all real 𝒂, 𝒃, 𝒄: (𝒂𝟐 + 𝒃𝟐) ≥ (𝒂 + 𝒃 + 𝒄)(𝒂 + 𝒃 − 𝒄)(𝒃 + 𝒄 − 𝒂)(𝒄 + 𝒂 − 𝒃)
Expanding both sides and moving to the left gives us:
2𝑎4 + 2𝑏 4 + 𝑐 4 − 2𝑎2 𝑐 2 − 2𝑏 2 𝑐 2 ≥ 0
This requires some ingenuity to factorise. Looking at the terms, we might try
(𝑎2 + 𝑏 2 − 𝑐 2 )2 , which gives us 𝑎4 + 𝑏 4 + 𝑐 4 + 2𝑎2 𝑏 2 − 2𝑎2 𝑐 2 − 2𝑏 2 𝑐 2 . Conveniently, this just
leaves 𝑎4 + 𝑏 4 − 2𝑎2 𝑏 2 left to make up to obtain the above expression. This factorises to (𝑎2 − 𝑏 2 )2 .
Thus we have (𝑎2 + 𝑏 2 − 𝑐 2 )2 + (𝑎2 − 𝑏 2 )2 ≥ 0. This is true by the Trivial Inequality.
8.
Let 𝒂, 𝒃, 𝒄, 𝒅 be positive real numbers which add to 1. Prove that:
𝒂𝟐
𝒃𝟐
𝒄𝟐
𝒅𝟐
𝟏
+
+
+
≥
𝒂+𝒃 𝒃+𝒄 𝒄+𝒅 𝒅+𝒂 𝟐
By the Cauchy-Schwarz Inequality:
𝑎2
𝑏2
𝑐2
𝑑2
((𝑎 + 𝑏) + (𝑏 + 𝑐) + (𝑐 + 𝑑) + (𝑑 + 𝑎)) (
+
+
+
) ≥ (𝑎 + 𝑏 + 𝑐 + 𝑑)2
𝑎+𝑏 𝑏+𝑐 𝑐+𝑑 𝑑+𝑎
𝑎2
𝑏2
𝑐2
𝑑2
2(𝑎 + 𝑏 + 𝑐 + 𝑑) (
+
+
+
) ≥ (𝑎 + 𝑏 + 𝑐 + 𝑑)2
𝑎+𝑏 𝑏+𝑐 𝑐+𝑑 𝑑+𝑎
(𝑎 + 𝑏 + 𝑐 + 𝑑)2
𝑎2
𝑏2
𝑐2
𝑑2
+
+
+
≥
𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑑 𝑑 + 𝑎 2(𝑎 + 𝑏 + 𝑐 + 𝑑)
𝑎2
𝑏2
𝑐2
𝑑2
1
+
+
+
≥
𝑎+𝑏 𝑏+𝑐 𝑐+𝑑 𝑑+𝑎 2
The last line follows because 𝑎 + 𝑏 + 𝑐 + 𝑑 = 1.
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