Prove that $\frac{x}{\sqrt{x-1}} \ge 2$ This is a simple inequality problem that I have found in a book (as part of a solution of a larger problem), however I have failed to prove it.
The only proof provided is that it is true since $(x-2)^2\ge0$ (I hope this could help you).
Can anyone please provide me with a proof or even the start of it.
 A: Note that for every $ x > 1 $, we have
\begin{align}
       \frac{x}{\sqrt{x - 1}}
& =    \frac{x - 1}{\sqrt{x - 1}} + \frac{1}{\sqrt{x - 1}} \\
& =    \sqrt{x - 1} + \frac{1}{\sqrt{x - 1}} \\
& \geq 2 \sqrt{\sqrt{x - 1} \cdot \frac{1}{\sqrt{x - 1}}} \qquad
       (\text{By the AM-GM Inequality.}) \\
& =    2.
\end{align}
A: An elementary approach: squares are positive so $(x-2)^2\ge 0$. Dividing by $4x^2$ you obtain
$$\frac{x^2-4x+4}{4x^2}\ge 0$$
Separating the fractions, this becomes
$$\frac{1}{4}\ge\frac{1}{x}-\frac{1}{x^2}$$
Since both quantities are positive if $x>1$, we can take the square root and 
$$\frac{1}{2}\ge\sqrt{\frac{1}{x}-\frac{1}{x^2}}=\frac{\sqrt{x-1}}{x}$$
Taking reciprocals yields the result:
$$\frac{x}{\sqrt{x-1}}\ge 2$$
A: A solution which uses the hint, keep in mind, everything is positive and we have $x>1$:
We start with
\begin{align}
\frac{x}{\sqrt{x-1}} \ge 2&\iff x\ge2{\sqrt{x-1}}\\
&\iff x^2\ge4(x-1)\\
&\iff x^2-4x+4\ge0
\end{align}
and now we see that
$$
x^2-4x+4=(x-2)^2\ge0 \text{ (hint)}
$$
and therefore we have
$$
\frac{x}{\sqrt{x-1}} \ge 2\iff(x-2)^2\ge0 
$$
which is true and therefore proves the above statement.
A: Hint: differentiate $\frac{x}{\sqrt{x-1}}$, find the minimum (which is $2$ at $x=2$). 
A: Note that
$$\frac{x}{\sqrt{x-1}} = \frac{t^2+1}{t} = t+\frac{1}{t} \ge 2$$
where $t = \sqrt{x-1} \ge 0$.
