RMO inequality problem 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?
 A: Note that we can rewrite your expression as
$$ \left( \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} \right) + \left( \frac{d^2}{e-1} + \frac{e^2}{d-1} \right) \geq 20. $$
Let us consider first the baby problem of finding the minimum value of $f(x) = \frac{x^2}{x-1}$ assuming $x > 1$. The minimum is readily seen to be attained at $x = 2$ with value $4$. 
Next, consider the problem of finding the minimum value of
$$ \frac{d^2}{e-1} + \frac{e^2}{d-1} =  f(d) \frac{d-1}{e-1} + f(e) \frac{e-1}{d-1} $$
assuming $d,e > 1$. The arithmetic-geometric inequality shows that
$$ f(d) \frac{d-1}{e-1} + f(e) \frac{e-1}{d-1} \geq 2 \sqrt{f(d)f(e)} \geq 2f(2) = 8 $$
where the minimum value of $8$ is indeed attained when $d = e = 2$.
Finally, consider the problem of finding the minimum value of
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} = f(a)\frac{a-1}{b-1} + f(b)\frac{b-1}{c-1} + f(c)\frac{c-1}{a-1} $$
assuming $a,b,c > 1$. Again, the arithmetic-geometric inequality shows that
$$ f(a)\frac{a-1}{b-1} + f(b)\frac{b-1}{c-1} + f(c)\frac{c-1}{a-1} \geq 3 (f(a)f(b)f(c))^{\frac{1}{3}} \geq 3f(2) = 12 $$
where the minimum indeed attained at $a = b = c = 2$.
Combining everything above, we obtain the required result.
A: 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 $a^2,b^2,c^2,d^2,e^2$
A: The functions $x^2$ and $(x-1)$ are both increasing in $x$.  Therefore, by the rearrangement inequality, the sum is $ \geq \sum \frac{a^2}{a-1}$ which is easier to control, in fact each term of this sum is $\geq 4$.
A: By Cauchy Schwarz, $($given expression$)(a+b+c+d+e-5)\geq(a+b+c+d+e)^2$
Now this shows $($given expression$)\geq \dfrac{(a+b+c+d+e)^2}{a+b+c+d+e-5}$
Now $(a+b+c+d+e)^2=(a+b+c+d+e-5)^2+2\times(a+b+c+d+e-5)\times 5+25$ hence on division by $a+b+c+d+e-5$ we get $a+b+c+d+e-5+10+\dfrac{25}{a+b+c+d+e-5}=(a+b+c+d+e-5)+\dfrac{25}{a+b+c+d+e-5}+10$
Letting $x=a+b+c+d+e-5$ we note $x\geq0$ and the above is $x+\dfrac{25}{x}+10\geq 10+10=20$ by AM-GM.
A: Note that $x^2-4(x-1) = (x-2)^2\ge 0$ for all $x\in\mathbb{R}$, and hence $x^2\ge 4(x-1)$ for all $x\in\mathbb{R}$. As such, we have
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 4\left(\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\right). $$
So it suffices to show
$$\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\ge 5.$$
Can you see how to finish?
