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  • Find the minimum value of $y = \frac{x^2}{x-9}$ using AM-GM inequality when $x \gt 9$.

How I tried to solve it:

Setting $p=x-9$, I did the following: $$y=\frac{(p+9)^2}{p}$$ $$y=\frac{p^2+18p+81}{p}$$ $$y=p+18+\frac{81}{p}$$ Since $x \gt 9$, $p\gt0$.

Using the AM-GM inequality, we can say, $$\frac{p+18+\frac{81}{p}}{3} \ge \sqrt[3]{p \cdot 18 \cdot \frac{81}{p}}$$ $$y \ge 27\sqrt[3]{2}$$ The actual minimum of $y$ is 36. Now, $y \gt 27\sqrt[3]{2}$ is certainly true, but it is surely not a satisfactory answer. This link here hints at the proper way to solve the problem.

How should the inequality be used in general so that it does not cause problems like this? Why is my solution inappropriate?

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  • $\begingroup$ You also need to find a value of $y$ that achieves your proposed minimum. Otherwise you just have an inequality. $\endgroup$ – Cheerful Parsnip Apr 20 '18 at 16:58
  • $\begingroup$ AM/GM is only optimal when all the addends have the same value. It's impossible that $p$, $18$ and $81/p$ are all equal. $\endgroup$ – Lord Shark the Unknown Apr 20 '18 at 17:13
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One has $$p+\frac{81}p\ge18$$ by AM/GM, with equality when $p=9$, so $$p+18+\frac{81}p\ge36,$$ with equality when $p=9$.

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By AM-GM $$\frac{x^2}{x-9}=\frac{x^2-81+81}{x-9}=x-9+\frac{81}{x-9}+18\geq2\sqrt{(x-9)\cdot\frac{81}{x-9}}+18=36.$$ The equality occurs for $x-9=\frac{81}{x-9},$ which says that $36$ is a minimal value.

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