Question

Given real numbers $a,b,c,d,e>1$ prove that
$\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\ge$ 20

Answer 1

Since $(a-2{)}^2\ge 0$
$\therefore a^2\ge 4(a-1)$
Since a > 1, we have $\frac{a^2}{a-1}\ge$4.
By applying AM-GM inequality, we get
$\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\ge 5\sqrt[5]{5\frac{a^2{b}^2{c}^2{d}^2{e}^2}{(a-1)(b-1)(c-1)(d-1)(e-1)}\ge }$ 20