Find the minimum value of $\frac{x^2}{x-1}$ for $x > 1.$

Question

Find the minimum value of $$\frac{x^2}{x-1}$$ for $x > 1.$

I can't use calculus, and I think the question is meant to be solved using the Trivial Inequality, the Mean Chain, and/or the Cauchy-Shwarz Inequality.

Any help would be greatly appreciated. Thanks!

Answer 1

Okay. Write: $\dfrac{x^2}{x-1} = x-1 + \dfrac{1}{x-1}+2\ge 3\sqrt[3]{2}$ by AM-GM inequality. Can you finish it?

Answer 2

Minimizing ${x^2\over x-1}$ for $x\gt1$ is the same as minimizing ${(u+1)^2\over u}$ for $u\gt0$. Since

$${(u+1)^2\over u}={u^2+2u+1\over u}=u+2+{1\over u}$$

it suffices to minimize $u+{1\over u}$ for $u\gt0$ and then add $2$. The arithmetic-geometric mean inequality says

$${1\over2}\left(u+{1\over u}\right)\ge\sqrt{u\cdot{1\over u}}=1$$

from which it's easy to see that the minimum occurs when $u={1\over u}=1$, which says the minimum of ${x^2\over x-1}$ occurs at $x=2$, with value ${2^2\over2-1}=4$.

If you don't already have AGM in your toolkit, you can prove what you need near by noting that for $u\gt0$

$$u+2+{1\over u}=4+u-2+{1\over u}=4+\left(\sqrt u-\sqrt{1\over u}\right)^2$$

which is clearly minimized when $\sqrt u={1\over\sqrt u}$.