<s>[INST] proving $ \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}\geq 20$ If $a,b,c,d,e>1$, Then prove that $\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq 20$
$\bf{My\; Try}::$ Using Cauchy- Schtwartz Inequality
$\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq \frac{(a+b+c+d+e)^2}{(c+d+e+a+b)-5}$
Now I did not understand how can i solve it
Help Required
Thanks
 [/INST] Apply CS inequality to get 
$$LHS > \frac{(a+b+c+d+e)^2}{a+b+c+d+e - 5} = M$$
Then let $y = a + b + c + d + e - 5$, then $y > 0$, and so
$$M = \frac{(y + 5)^2}{y}= 10 + y + \frac{25}{y} \ge 10  + 2\sqrt{25} \overset{\strut\text{AM-GM}}= 10 + 10 = 20$$
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