Trivial Inequality

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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.



  • Show that $x^2+y^4\geq 2x+4y^2-4$ for all real $x$.
  • 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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