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?

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.
