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

Question

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!

Answer 1

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 $x =\sqrt {2 \pm \sqrt{3}}$

Answer 2

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 .

Answer 3

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

Answer 4

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 $z>0$