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.

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 $a=b$.

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 $x=0$. $_\square$

Show that $x^2 + 2x + 2 > 0$ for all real values of $x$.

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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 $x$.

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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 $x$.

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We have $\begin{aligned} \dfrac{4x^2 + 12x + 10}{9x^2 - 42x + 50} & = \dfrac{4x^2 + 12x + 9 + 1}{9x^2 - 42x + 49 + 1}\\ \\ & = \dfrac{{(2x + 3)}^2 + 1}{{(3x - 7)}^2 + 1}. \end{aligned}$ 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, $\frac{{(2x + 3)}^2 + 1}{{(3x - 7)}^2 + 1} > 0$. $_\square$