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?
 A: For convenience let $y=x-1$, so that the expression becomes $$\frac{(y+1)^2}y=\frac{y^2+2y+1}y=y+2+\frac1y\;,$$ where $y>0$. Clearly this attains its minimum when $f(y)=y+\frac1y$ attains its minimum. Notice that $$f\left(\frac1y\right)=\frac1y+\frac1{1/y}=\frac1y+y=f(y)\;,$$ so that $f$ has a kind of symmetry around $y=1$. In particular, if $f$ is increasing for $y>1$, then it’s decreasing for $0<y<1$ and has a minimum at $y=1$. Suppose that $1<y<z$; then
$$\begin{align*}
f(z)-f(y)&=\left(z+\frac1z\right)-\left(y+\frac1y\right)\\
&=(z-y)-\left(\frac1y-\frac1z\right)\\
&=(z-y)-\frac{z-y}{yz}\\
&=(z-y)\left(1-\frac1{yz}\right)\;,
\end{align*}$$
which shows that $f$ is indeed increasing for $y>1$. (Why?) Thus, $f$ has a minimum at $y=1$, and the original function has a minimum at ... ?
A: 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.
A: Take the derivative and set it equal to 0.  Denote the $x$ that satisfies this as $x_0$.  Then plug in the value $x=1$.  Is $f(1)<f(x_0)$?  The minimum is $\min(f(1),f(x_0))$.
A: You could use the second derivative test.
$f(x)=\frac{x^2}{x-1}$
$f'(x)=\frac{x(x-2)}{(x-1)^2}$
so $f'(x)=0$ for $x=0,2\,\,\,\,\,\,\,\,\,\,$ So extremum at x=0 and 2
$f''(2)>0$
So minima at $x=2$
A: The standard method for such a problem is calculus as used by @Swapnil Tripathi. If you want to avoid it:
Instead of asking for the minimum of 
$$f(x)=\frac{x^2}{x-1}, \forall x > 1$$ 
you can ask for the maximum of 
$$\frac{1}{f(x)}=\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}, \forall 0<x<1$$
 So the maximum is at 
$$\frac{1}{x}-\frac{1}{2}=0$$
and so 
$$x=2$$
A: You can try plotting it and just look for the minimum:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2F%28x-1%29+for+x%3E1
Look at the second plot it provides (which has $x>1$). I actually don't remember precalc techniques to find minima - I presume you can't just take a first order condition, because that would be calculus?
