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I have just learned about the AM - GM inequality and I have done a couple of normal problems. However, I'm stuck in this one.. I feel like I have to rewrite it as a summation but I don't see how.

What is the minimum value of $\frac{x^2}{(x-9)}$ for $x>9$

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Hint:

$$\frac{x^2}{x-9}=x-9+\frac{9(2x-9)}{x-9}$$

Now apply AM-GM for $x-9$ and $\frac{18x-81}{x-9}$.

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Hint:

$$\frac{x^2}{x-9}=x+9+\frac{81}{x-9}$$

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Your question can be re-written as:
Find out the minimum value of $\frac{(t+9)^2}{t}, t>0$. Re-arranging, $t+\frac{81}{t} + 18$. Now, apply AM-GM inequality.

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