Given real numbers $a,b,c,d,e>1$. Prove that \[ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge 20 \]
Problem
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Tags: cyclic inequality
12.12.2013 18:55
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 $a=b=c=d=e=2$.
14.12.2013 03:26
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} $ $\ge 5\sqrt[5]{\frac{4(a-1)}{c-1}\cdot \frac{4(b-1)}{d-1}\cdot \frac{4(c-1)}{e-1}\cdot \frac{4(d-1)}{a-1}\cdot \frac{4(e-1)}{b-1} }= 20 .$
22.08.2015 19:44
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.
23.09.2015 20:52
Just use $AM-GM$ and the trivial inequality $(a-2)^2>=0$. This solves the problem.....
04.12.2015 11:45
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. $\Square$
24.07.2017 15:50
Very similar question asked in Russia 1992 paper. Just 2 variables instead of 5.
04.10.2017 18:52
So the lower bound of a,B,c,d,e is an extra information?
04.10.2017 19:03
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?
04.10.2017 19:25
Since the last result is the trivial inequality hence it is obviously true @Mdz02
04.10.2017 19:36
MdZ02 wrote: Hey! How do you say that its obvious at last? We are in $\mathbb{R}$. (Not $\mathbb{C}$)
05.10.2018 08:40
Can any one show briefly the steps of am- gm
30.09.2019 23:25
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.
03.10.2019 16:26
$$\frac{a^2}{c-1}+4(c-1)\ge 2\sqrt{\frac{a^2}{c-1}\cdot 4(c-1)}=4a$$Hence $$\sum_{cyc}\frac{a^2}{c-1}\ge \sum_{cyc}(4a-4c+4)=4\cdot 5=20$$
07.10.2020 10:55
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)
05.11.2020 13:58
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