TITLE: RMO inequality problem
QUESTION [7 upvotes]: If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?

REPLY [8 votes]: Using Cauchy-Scwarz to 
\begin{align*}
\frac{a_1}{\sqrt{x_1}}, \frac{a_2}{\sqrt{x_2}}, \ldots, \frac{a_n}{\sqrt{x_n}} \\
\sqrt{x_1}, \sqrt{x_2}, \ldots, \sqrt{x_n}
\end{align*}
we get
\begin{align*}
\frac{a_1^2}{x_1}+\frac{a_2^2}{x_2}+\cdots + \frac{a_n^2}{x_n} \geq \frac{(a_1+a_2+\cdots + a_n)^2}{x_1+x_2+\cdots+x_n}
\end{align*}
Hence we have
\begin{align*}
\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{d^2}{e-1}+\frac{c^2}{a-1}+\frac{e^2}{d-1} \geq \frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}
\end{align*}
Putting $x=a+b+c+d+e$, we need to prove
\begin{align*}
\frac{x^2}{x-5} \geq 20
\end{align*}Since $x > 5$, this can be written as  $x^2 -20x + 100 \geq 0$. This follows readily since 
\begin{align*} x^2-20x+100 = (x-10)^2\end{align*}
The proof also shows that the numerators can be any permutation of $a^2,b^2,c^2,d^2,e^2$