Trivial Inequality

The trivial inequality is an inequality that states that the square of any real number is nonnegative. Its name comes from its simplicity and straightforwardness.

Statement

For all real numbers $x$ , ${x^2 \ge 0}$ .

Proof

We proceed by contradiction. Suppose there exists a real $x$ such that $x^2<0$ . We can have either $x=0$ , $x>0$ , or $x<0$ . If $x=0$ , then there is a clear contradiction, as $x^2 = 0^2 \not < 0$ . If $x>0$ , then $x^2 < 0$ gives $x < \frac{0}{x} = 0$ upon division by $x$ (which is positive), so this case also leads to a contradiction. Finally, if $x<0$ , then $x^2 < 0$ gives $x > \frac{0}{x} = 0$ upon division by $x$ (which is negative), and yet again we have a contradiction.

Therefore, $x^2 \ge 0$ for all real $x$ , as claimed.

Applications

The trivial inequality is one of the most commonly used theorems in mathematics. It is very well-known and does not require proof.

One application is maximizing and minimizing quadratic functions. It gives an easy proof of the two-variable case of the Arithmetic Mean-Geometric Mean inequality:

Suppose that $x$ and $y$ are nonnegative reals. By the trivial inequality, we have $(x-y)^2 \geq 0$ , or $x^2-2xy+y^2 \geq 0$ . Adding $4xy$ to both sides, we get $x^2+2xy+y^2 = (x+y)^2 \geq 4xy$ . Since both sides of the inequality are nonnegative, it is equivalent to $x+y \ge 2\sqrt{xy}$ , and thus we have $\frac{x+y}{2} \geq \sqrt{xy},$ as desired.

Problems

Find all integer solutions $x,y,z$ of the equation $x^2+5y^2+10z^2=2z+6yz+4xy-1$ .
(AIME 1992) Triangle $ABC$ has $AB$ $=9$ and $BC: AC=40: 41$ . What is the largest area that this triangle can have?

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Trivial Inequality

Contents

Statement

Proof

Applications

Problems

See also