Proving algebraic inequalities.
Definition of algebraic inequality: An algebraic inequality is the inequality of two
algebraic expressions. For example: 2 x  6  0, x 2  6  3x, a 2  b 2  2ab.
Difference between proving and solving inequalities.
To solve an inequality is to find all the solutions of that inequality.
Example: Solve the inequality 2x  6  0.
Solution: For solving this inequality we transform it into an equivalent inequality for
which is easier to determine the solutions. Two inequalities are said to be equivalent if
the share the same set of solutions. The process of solving that inequality consists of the
following steps:
2x  6  0
2 x  6
x  6 / 2
x  3
Obviously, the set of solutions of the last inequality is the interval (,  3).
To prove an inequality is to determine whether the inequality is always true for all the
values of the variables on a certain set of numbers.
Example: Prove that x( x  1)  0 , for all positive values of x.
Solution:
As x is positive, then x  0 (*). As x  0 , then x  1  1  0 . That is
x  1  0 (**). Now, using (*) and (**), we have that x( x  1)  0 .
Properties of inequalities:
Now we present some properties that can be use to prove inequalities. Let us assume that
a, b, c, a1 , b1 , c1 , a2 , a3 ,...an are arbitrary real numbers. (Note: The symbol "" means
“implies”)
1. a  b  a  c  a  b
2. (a  b and b  c)  a  c
3. (a  b and c  0)  ac  bc
4. (a  b  0 and a1  b1  0)  aa1  bb1
5. (a  b  0 and n  N )  a n  b n
6. (a  b  0 and n  N )  a1 / n  b1 / n
7. a 2  0
2
2
2
2
2
8. a1  a2  a3  a4  ...  an  0
The properties 1 through 6 are true when you replace the symbol "" by "" .
Examples of proving inequalities.
1. Prove that for all real numbers a, b , it is valid the inequality a 2  b 2  2ab.
Solution.
Analysis: You should analyze the inequality and notice that it is equivalent to
2
a  b 2  2ab  0. In other hand a 2  b 2  2ab  (a  b) 2 and we know from the
property (7) the square of a number is always a number greater than or equal to zero.
Therefore, this suggests the following proof
Proof: For arbitrary real numbers a, b . We have the following sequence of
implications.
(a  b) 2  0  a 2  b 2  2ab  0  a 2  b 2  2ab.
2. Prove that for all positive number a, it is valid that c  1/ c  2.
Solution:
Analysis: You can find a relation between this inequality and the Example 1. As
a is positive it can be represented as ( c ) 2 and 1 / c can be represented as
(1 / c ) 2 . At the same time, we have that 2(c)(1 / c )  2 . That is why we get the
following idea for proving this inequality.
Proof: We start assuming that the inequality a 2  b 2  2ab has been already
proved (see example 1). In that inequality let do the substitutions a  c and
b  1 / c . Therefore, we have
( c ) 2  (1 / c ) 2  2( c )(1 / c )  c  1 / c  2
Exercises
1. Prove that a / b  b / a  2, if a and b are positive real numbers. Find the values
of a and b for which the equality is true.
2. Prove that a 2  (1 / 3)ab  b 2  0 ( Hint: complete squares and use property (8))
ab
 ab , if a and b are positive real numbers.
3. Show that
2
4. Find the all the values of t such that a 2  tab  b 2  0 .
5. Prove that a 2  b 2  c 2  ab  bc  ac for all real numbers a, b, and c.
6. Prove that x 2  4 y 2  1  2 xy  x  2 y for all real numbers x and y.
abcd 4
 abcd , is true for a, b, c, and d are positive real
7. Prove that
4
ab cd

abcd
2
2 and

numbers. ( Hint: Take into consideration that
4
2
exercise (3))
abc 3
 abc , is true for a, b, and c are positive real numbers. (
3
abc
.
Hint: Apply the result of the exercise (7) considering d 
3
9. Prove that (a 2  b 2 )(c 2  d 2 )  (ac  bd ) 2 for all real numbers a, b, c, and d.
8. Prove that