# Proving the inequality that $\dfrac{x^2 + x^{-2}}{x-x^{-1}} \geq 2 \sqrt{2}$ for $x > 1$

Question: Show that $$\dfrac{x^2 + x^{-2}}{x-x^{-1}} \geq 2 \sqrt{2}$$ for $$x > 1$$.

My attempts: After spending some time trying to prove it by $$AM-GM$$ and with algebraic manipulation, I tried to use trigonometric substitutions like letting $$x = \tan\theta$$ and $$x = \sin\theta$$ although I was still unsuccessful. I know that this can be proven with calculus, however I am looking to prove this without the aid of calculus. Any help would be appreciated!

• Write it as $(x-x^{-1})+{\frac{2}{(x-x^-1)}}$ then apply AM-GM Jun 4, 2020 at 14:19

Let $$a=x-x^{-1}$$. Then we want to prove $$\frac{a^2+2}a\ge 2\sqrt{2}, a>0$$ which rearranges to $$(a-\sqrt{2}) ^2\ge0$$.

If $$x =1$$, we have $$x-x^{-1} =0$$ so I assume $$x>1$$.

Put $$t = x- \frac1x$$. Then $$t>0$$ and $$x^2 + \frac{1}{x^2}=t^2 +2$$

Now $$\frac{x^2 + x^{-2}}{x - x^{-1}} = \frac{t^2+2}{t}=t+\frac2t\geq2\sqrt2$$ by AM-GM. The equality holds when $$t^2=2$$, or $$x =\sqrt {2 \pm \sqrt{3}}$$

Hint: use $$x = \sec \theta, \sec^2 \theta = 1+ \tan^2 \theta$$. Of course there are many steps you need to go through, and this gives you a direction on how to proceed .

Let $$x-\dfrac1x=y$$ to find $$\dfrac{y^2+2}y=z\text{(say)}$$

$$\implies y^2-zy+2=0$$

As $$z$$ is real, the discriminant $$\ge0\implies (-z)^2\ge4\cdot2\cdot1$$

Now $$x-\dfrac1x$$ will be $$>0$$ if $$x>\dfrac1x\iff x<-1$$ or $$>1$$

In that case $$z>0$$