<s>[INST] Min/Max Problem Find the minimum of $$\frac{x^2}{x-1}$$ for $x > 1$.
I tried using AM-GM and Cauchy-Schwarz, but after using them I ended up with $x>1$ which was what was given. How should I start?
 [/INST] The shape of $f(x)$ for $x>1$ is a steep drop from infinite heights at $x=1^+$ going dwon but womehow going up looking like $f(x)\approx x$ for very large $x$.  So there will be a minimum.
A cute way to find the minimum is to simplify the expression, relying on the fact that $x \neq 1$:
$$
\frac{x^2}{x-1} = \frac{x^2-x}{x-1} + \frac{x}{x-1} = \frac{x^2-x}{x-1} + \frac{x-1}{x-1}
+ \frac{1}{x-1} = x + 1 + \frac{1}{x-1}$$
Then if we had calculus, we would find the minimum by taking the derivative of $x + 1 + \frac{1}{x-1}$ and setting it to zero, getting $x=2, f(x) = 4$, but in pre-calc that is, well, post-course.  
We can find the minimum without calculus by rewriting 
$$
f(x) = 2 + (x-1) + \frac{1}{x-1} = 2 + v + \frac{1}{v}$$
with $v = x-1$.  Then you know that $v + \frac{1}{v}$ has a minimum at $v=1$.
Which means $f(x)$ is its minimum at $v=1 \Rightarrow x = 2$ as before.
By the way, we can see that $v+\frac{1}{v}$ is always at least $2$ by:
$$
v + \frac{1}{v} -2 = \left( \sqrt{v} - \sqrt{\frac{1}{v}} \right)^2 \geq 0
$$
Then where is $v + \frac{1}{v} = 2$? At $v=1$ of course.
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