Normal AM-GM inequality

Question

Minimum value of $\frac{x^2}{x-9}$

I was asked to use the AM-GM inequality to solve this, I was thinking that I should express it into 2 fractions, and so I did, I tried to express $x^2$ as $\frac{(x-9)^2 +18x -81 }{x-9}$, though I couldn't find an answer.

On which interval do you want to minimise this? – Angina Seng Oct 02 '17 at 16:16 — Angina Seng, Oct 02 '17 at 16:16
the Minimum is $$-\infty$$ – Dr. Sonnhard Graubner Oct 02 '17 at 16:19 — Dr. Sonnhard Graubner, Oct 02 '17 at 16:19
Please, you should use MathJax to format your posts by now. – gen-ℤ ready to perish Oct 02 '17 at 16:55 — gen-ℤ ready to perish, Oct 02 '17 at 16:55

score 2 · Answer 1 · answered Oct 02 '17 at 16:22

2

Hint:

Set $x-9=h$ $$\dfrac{x^2}{x-9}=\dfrac{(h+9)^2}h=h+\dfrac{9^2}h+18$$

If $h>0\iff x-9>0$ $$\dfrac{h+\dfrac{9^2}h}2\ge\sqrt{h\cdot\dfrac{9^2}h}=?$$

answered Oct 02 '17 at 16:22

lab bhattacharjee

274,582

score 0 · Answer 2 · answered May 29 '20 at 07:36

0

Well, I'll try $$\frac{x^2}{x-9}=\frac{x^2-81+81}{x-9} =x+9+\frac{81}{x-9}=x-9+\frac{81}{x-9}+18$$ $$x-9+\frac{81}{x-9}+18 \ge 2\sqrt{(x-9)\left(\frac{81}{x-9}\right)}+18=2\cdot9+18=36$$

answered May 29 '20 at 07:36

Anas A. Ibrahim

1,884
7
15

Normal AM-GM inequality

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