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.


For all real numbers $x$, $x^2 \ge 0$, equality holds if and only if $x = 0$.


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.


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.



  • 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


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


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