# Trivial Inequality

(Redirected from 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 $x$, $x^2 \ge 0$, equality holds if and only if $x = 0$.

## Proof

We can have either $x=0$, $x>0$, or $x<0$. If $x=0$, then $x^2 = 0^2 \ge 0$. If $x>0$, then $x^2 = (x)(x) > 0$ by the closure of the set of positive numbers under multiplication. Finally, if $x<0$, then $x^2 = (-x)(-x) > 0,$ again by the closure of the set of positive numbers under multiplication.

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

### Introductory

• Find all integer solutions $x,y,z$ of the equation $x^2+5y^2+10z^2=4xy+6yz+2z-1$.
• Show that $\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1$. Solution
• Show that $x^2+y^4\geq 2x+4y^2-4$ for all real $x$.

### Intermediate

• Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$. What is the largest area that this triangle can have? (AIME 1992)

• Let $c$ be the length of the hypotenuse of a right triangle whose two other sides have lengths $a$ and $b$. Prove that $a+b\le c\sqrt{2}$. When does the equality hold? (1969 Canadian MO)