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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

juantheron
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  • The problem with what you have is that you don't add fractions just by adding the numerators and then dividing by the sum of the denominators. Also, if you use the Cauchy-Swartz inequality the inequality sign should go the other way. – Mehta Dec 12 '13 at 03:43
  • You are nearly done. Divide num and den of RHS by (a+b+...) and argue what could be the minimum value of the fraction. – Macavity Dec 12 '13 at 03:49
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    You are done. For every $x>5$, $x^2/(x-5)\geq 20$. – Julien Dec 12 '13 at 03:50

1 Answers1

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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 $$M = \frac{(y + 5)^2}{y}= 10 + y + \frac{25}{y} \ge 10 + 2\sqrt{25} \overset{\strut\text{AM-GM}}= 10 + 10 = 20$$

chubakueno
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DeepSea
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    An alternative approach(not worthy of a complete answer): Let $a+b+c+d+e=5\sec^2(\alpha)$, and we have $LHS\ge \frac{20}{\sin^2(2\alpha)}\ge 20$ – chubakueno Dec 12 '13 at 18:34