1

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!

Quantum Pizza
  • 146
  • I tried the Mean Inequality Chain and Cauchy-Shwarz, but they aren't working. Maybe I'm using them in the wrong way. And no, I don't know anything about calculus, so no calculus :D. – Quantum Pizza May 12 '17 at 01:13
  • That's a good thing to state when asking the problem then! This problem can be tackled in multiple ways, so its good to know which way you want. Now I'll have to think about that way... – Bob Krueger May 12 '17 at 01:18
  • Okay, thanks. I'll edit the question to make it easier to understand. – Quantum Pizza May 12 '17 at 01:19
  • Related: https://math.stackexchange.com/questions/472169/find-extreme-values-of-frac2xx²4 – lab bhattacharjee May 12 '17 at 01:26

2 Answers2

5

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 $\sqrt u={1\over\sqrt u}$.

Barry Cipra
  • 79,832
  • Barely beat me to it! Another important note is why anyone would every think about changing this problem into maximizing $\frac{(u+1)^2}{u}$, in order to help someone learn how to solve these inequalities. (The change results from the sort-of-weirdness of the condition $x > 1$; we would expect $x > 0$.) – Bob Krueger May 12 '17 at 01:24
  • @Bob1123, sorry about that! Your comment, prompting the OP's clarification, really helped. – Barry Cipra May 12 '17 at 01:31
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?

DeepSea
  • 77,651
  • 1
    Quite honestly, I don't see how establishing $3\sqrt[3]2\approx3.77976$ helps identify $4$ as the actual minimum value. – Barry Cipra May 12 '17 at 01:30
  • @BarryCipra: The point here is not really finding the exact amount of the minimum, rather the trick involved and how one could rewrite the rational expression to reveal some insight or connection to the main idea which is the use of AM-GM or CS inequalities w/o making it way too long ... – DeepSea May 12 '17 at 01:32
  • OK, but the problem is to find the minimum, and I really don't see how to finish things off from the inequality you stated. It looks like a dead end to me. Obviously the OP has found your answer helpful, though, so so be it. – Barry Cipra May 12 '17 at 01:42
  • Yes, your answer is correct and perhaps better than mine. No contest. Here OP wants to have fun and wants to see "different" strategy.... – DeepSea May 12 '17 at 01:44
  • Ah, how about writing ${x^2\over x-1}=(x-1)+{1\over x-1}+1+1\ge\ldots$! – Barry Cipra May 12 '17 at 01:50