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Algebra • Topic 3

Algebraic Inequalities

Before tackling advanced inequalities like AM-GM or Cauchy-Schwarz, one must master the fundamental algebraic manipulations, particularly the "Trivial Inequality."

The Trivial Inequality

Theorem (Trivial Inequality)

For any real number

x

x^{2} \geq 0

Equality holds if and only if

x = 0

This simple fact is the foundation of the Sum of Squares (SOS) method.

Immediate Consequences

Square of Difference: $(a - b)^{2} \geq 0 ⟹ a^{2} - 2 ab + b^{2} \geq 0 ⟹ a^{2} + b^{2} \geq 2 ab$ .
Reciprocals: For $x > 0$ , $x + \frac{1}{x} \geq 2$ . (Set $a = x, b = 1/ x$ ).

Basic Techniques

1. Completing the Square

If you can write an expression

E

E = A^{2} + B^{2} + C

, then the minimum value of

E

C

(achieved when

A = 0

and

B = 0

Example

Find the minimum value of

2 x^{2} - 4 x + 7

Proof (Solution).

2 (x^{2} - 2 x) + 7 = 2 (x^{2} - 2 x + 1 - 1) + 7 = 2 (x - 1)^{2} - 2 + 7 = 2 (x - 1)^{2} + 5

Since

(x - 1)^{2} \geq 0

, the minimum is 5, occurring at

x = 1

. ∎

2. Titu's Lemma (Introduction)

A useful variant of Cauchy-Schwarz that can be proved using simple algebra for small cases.

\frac{a ^{2}}{x} + \frac{b ^{2}}{y} \geq \frac{( a + b ) ^{2}}{x + y}

Proof. This is equivalent to $(a^{2} y + b^{2} x) (x + y) \geq (a + b)^{2} x y$ . Expanding and simplifying leads to $(a y - b x)^{2} \geq 0$ , which is true. ∎

Chain Inequalities

Many problems involve proving

A \geq B \geq C

. Often, you prove

A \geq B

and

B \geq C

separately.

Practice Problems

Exercise (Problem 1)

Prove that for all real numbers

x, y, z

x^{2} + y^{2} + z^{2} \geq x y + yz + z x

(Hint: Multiply by 2 and look for squares of differences)

Exercise (Problem 2)

Prove that for

a, b > 0

\frac{a}{b} + \frac{b}{a} \geq 2

Exercise (Problem 3)

Determine the minimum value of

x^{2} + y^{2} - 4 x + 6 y + 15

for real

x, y