# Trivial Inequality

The trivial inequality simply states that squares of real numbers are always non-negative. That is, for any real number \(x\), we have \(x^2\ge 0\). Although this is trivial to see, it gives birth to many non-trivial and useful inequalities, and is often very useful in solving Olympiad inequality problems.

#### Contents

## Definition

We start by learning the definition and then its relationship with other non-trivial inequality.

The trivial inequality states that for any real number \(x\), we have

\[ x^2 \geq 0\]

and equality holds if and only if \(x = 0\).

To get a feel on how it produces non-trivial results, we set \(x=\sqrt a - \sqrt b\) for some non-negative reals \(a,b\). Then by the trivial inequality we have

\[x^2\ge 0 \implies \left(\sqrt a - \sqrt b\right)^2\ge 0\implies a+b-2\sqrt{ab}\ge 0\implies \dfrac{a+b}{2}\ge \sqrt{ab},\]

which is the two variable case of AM-GM inequality and is very useful in solving many Olympiad level problems. Of course, the equality occurs when \(x=0\), that is \(\sqrt a -\sqrt b =0\), which simplifies to \(a=b\).

## Proof

In this section we present a short proof for the trivial inequality.

For any \(x\in\mathbb{R}\), we can have three cases. Either it's positive, negative, or \(0\). We can rewrite any negative \(x\) as \(-y\) for some positive \(y\in\mathbb{R}\). A positive number multiplied by a positive number is positive.

So if \(x\) is positive, we have \(x^2=x\times x > 0\).

If \(x\) is negative, we substitute \(x=-y\). Now \(y\times y > 0\). Then

\[x^2=(-y)^2=(-1)^2 \times y^2 = 1 \times y\times y > 0.\]

If \(x=0\), then \(x^2=x\times x=0\times 0=0\).

Hence for any real \(x\in\mathbb{R}\) we have \(x^2\ge 0\), and equality holds if and only if \(x=0\). \(_\square\)

## Problem Solving

## Show that \( x^2 + 2x + 2 > 0 \) for all real values of \( x \).

Rewrite the left side of the inequality as \[ x^2 + 2x + 2 = x^2 + 2x + 1 + 1 = (x+1) ^2 + 1 \] by completing the square. By the trivial inequality, we have \((x+1)^2\ge 0\), thus \((x+1)^2+1\ge 1>0\). \(_\square\)

## Show that \( x + \frac1x \ge 2 \) for all positive real values of \( x \).

Subtract 2 from both sides to get \( x -2 + \frac1x \ge 0 \). Since \(x\) is positive, we can multiply both sides by \(x,\) making the inequality \( x^2 -2x + 1 = (x-1)^2 \ge 0,\) which is true by the trivial inequality. \(_\square\)

Show that \(\frac{4x^2 + 12x + 10}{9x^2 + 42x + 50} > 0\) for all real values of \(x\).

We have \[\begin{align}

\dfrac{4x^2 + 12x + 10}{9x^2 + 42x + 50} & = \dfrac{4x^2 + 12x + 9 + 1}{9x^2 - 42x + 49 + 1}\\ \\ & = \dfrac{{(2x + 3)}^2 + 1}{{(3x - 7)}^2 + 1}. \end{align}\] Now, the range of any square is \([0,\infty)\). In other words, all squares when simplified give values \(\geq 0\). So, \({(2x + 3)}^2\) and \({(3x + 7)}^2\) each give values \(\geq 0\). So, \({(2x + 3)}^2 + 1\) and \({(3x + 7)}^2 + 1\) each give values \(\geq 1\). So, \(\frac{{(2x + 3)}^2 + 1}{{(3x - 7)}^2 + 1} > 0\). \(_\square\)

**Cite as:**Trivial Inequality.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/trivial-inequality/