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!
Best Answer
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 $\sqrt u={1\over\sqrt u}$.