By what is also known as Titu's form of Cauchy-Schwarz we have $\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \dfrac {(\sum a)^2} {\sum a - 5} \ge 20$, since it comes to $(\sum a-10)^2 \ge 0$. Equality occurs for $a=b=c=d=e=2$.

By AM--GM ,we have $\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}$ $\ge\frac{4(a-1)}{c-1}+\frac{4(b-1)}{d-1}+\frac{4(c-1)}{e-1}+\frac{4(d-1)}{a-1}+\frac{4(e-1)}{b-1} $ $\ge 5\sqrt[5]{\frac{4(a-1)}{c-1}\cdot \frac{4(b-1)}{d-1}\cdot \frac{4(c-1)}{e-1}\cdot \frac{4(d-1)}{a-1}\cdot \frac{4(e-1)}{b-1} }= 20 .$

We have by C-S, $ \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}{a+b+c+d+e-5}$. Thus it suffices to show that, if $a+b+...+e=k$, that $k^2 - 20k + 100 \geq 0 \Longleftrightarrow (k-10)^2 \geq 0$ which is obviously true.

Just use $AM-GM$ and the trivial inequality $(a-2)^2>=0$. This solves the problem.....

My Solution. Let us try to apply Engel Form of CS inequality, or the Titu's Lemma on LHS. We will get $-$ $$ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \frac{(a+b+c+d+e)^2}{(a-1)+(b-1)+(c-1)+(d-1)+(e-1)}=\frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}.$$Let us denote $S=a+b+c+d+e.$ So it remains to prove that $-$ $$\frac{S^2}{S-5}\ge 20 \Leftrightarrow S^2\ge 20S+100\Leftrightarrow S^2-20S+100\ge 0\Leftrightarrow (S-10)^2\ge 0,$$which is obvious. $\Square$

Very similar question asked in Russia 1992 paper. Just 2 variables instead of 5.

So the lower bound of a,B,c,d,e is an extra information?

eshan wrote: My Solution. Let us try to apply Engel Form of CS inequality, or the Titu's Lemma on LHS. We will get $-$ $$ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \frac{(a+b+c+d+e)^2}{(a-1)+(b-1)+(c-1)+(d-1)+(e-1)}=\frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}.$$Let us denote $S=a+b+c+d+e.$ So it remains to prove that $-$ $$\frac{S^2}{S-5}\ge 20 \Leftrightarrow S^2\ge 20S+100\Leftrightarrow S^2-20S+100\ge 0\Leftrightarrow (S-10)^2\ge 0,$$which is obvious. $\Square$ Hey! How do you say that its obvious at last?

Since the last result is the trivial inequality hence it is obviously true @Mdz02

MdZ02 wrote: Hey! How do you say that its obvious at last? We are in $\mathbb{R}$. (Not $\mathbb{C}$)

Can any one show briefly the steps of am- gm

Akshaykumar wrote: Can any one show briefly the steps of am- gm Apply AM-GM on $(a/2)^2$ and $1$. You will get $[{(a/2) ^2}+1]/2>=a/2$ Then solve according to squing's solution.

$$\frac{a^2}{c-1}+4(c-1)\ge 2\sqrt{\frac{a^2}{c-1}\cdot 4(c-1)}=4a$$Hence $$\sum_{cyc}\frac{a^2}{c-1}\ge \sum_{cyc}(4a-4c+4)=4\cdot 5=20$$

eshan wrote: My Solution. Let us try to apply Engel Form of CS inequality, or the Titu's Lemma on LHS. We will get $-$ $$ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \frac{(a+b+c+d+e)^2}{(a-1)+(b-1)+(c-1)+(d-1)+(e-1)}=\frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}.$$Let us denote $S=a+b+c+d+e.$ So it remains to prove that $-$ $$\frac{S^2}{S-5}\ge 20 \Leftrightarrow S^2\ge 20S+100\Leftrightarrow S^2-20S+100\ge 0\Leftrightarrow (S-10)^2\ge 0,$$which is obvious. $\Square$ it is very important to mention that S > 5, since you can mulitply or divide both sides of a inequality by a positive number only (so the condition that a,b,c,d,e > 1 is indeed very important)

Raj2000 wrote: So the lower bound of a,B,c,d,e is an extra information? The lower bound was necessary because then a-1,b-1,c-1,d-1,e-1 are positive reals and you can use Titu or AM GM if the variables are positive reals