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 \(a=b=c=d=e=2\)

Method 2:

We know that \((a-2)^2 \ge 0\) so \(a^2 \ge 4(a-1)\). Similarly we get \( \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \frac{4(a-1)}{c-1}+\frac{4(b-1)}{d-1}+\frac{4(c-1)}{e-1}+\frac{4(d-1)}{a-1}+\frac{4(e-1)}{b-1} \\ \ge 5\sqrt[5]{\frac{4(a-1)}{c-1}\cdot \frac{4(b-1)}{d-1}\cdot \frac{4(c-1)}{e-1}\cdot \frac{4(d-1)}{a-1}\cdot \frac{4(e-1)}{b-1} }= 20 . \)

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## Comments

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TopNewest@Rajdeep Dhingra

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You could tell me those problems and I may be able to help you ? See this

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http://olympiads.hbcse.tifr.res.in/uploads/crmo-2013-paper-4 here question number 3

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Method 1:

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 \(a=b=c=d=e=2\)

Method 2:

We know that \((a-2)^2 \ge 0\) so \(a^2 \ge 4(a-1)\). Similarly we get \( \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge \frac{4(a-1)}{c-1}+\frac{4(b-1)}{d-1}+\frac{4(c-1)}{e-1}+\frac{4(d-1)}{a-1}+\frac{4(e-1)}{b-1} \\ \ge 5\sqrt[5]{\frac{4(a-1)}{c-1}\cdot \frac{4(b-1)}{d-1}\cdot \frac{4(c-1)}{e-1}\cdot \frac{4(d-1)}{a-1}\cdot \frac{4(e-1)}{b-1} }= 20 . \)

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add me also in ur hangouts ... I am also preparing for RMO! I'll give you a hand in solving the problems !!

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Give me your Email ID.

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hrithik.nambiar2002@gmail.com

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thanks!

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Which class do ya study in ?

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u can use any of the following books 1 . arthur engel 2. pre college mathematics 3 . rmo and inmo prep booklet by rajeev manocha

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Can you guys help me in solving my note named :"Primes filled with primes".

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