# KVS RMO 2009 Q2

I spent around an hour to solve this one:

Let $$a, b, c, d, e$$ be positive real numbers. Prove that

$\large\ \frac { a }{ b+c } +\frac { b }{ c+d } +\frac { c }{ d+e } +\frac { d }{ e+a } +\frac { e }{ a+b } \ge \frac { 5 }{ 2 }$

I did this by applying Titu's lemma:

$\large\ \frac { a }{ b+c } +\frac { b }{ c+d } +\frac { c }{ d+e } +\frac { d }{ e+a } +\frac { e }{ a+b } =\frac { { a }^{ 2 } }{ ab+ac } +\frac { { b }^{ 2 } }{ bc+bd } +\frac { { c }^{ 2 } }{ cd+ce } +\frac { { d }^{ 2 } }{ de+ad } +\frac { { e }^{ 2 } }{ ae+be } \ge \frac { { (a+b+c+d+e })^{ 2 } }{ \sum { ab } }$

Because $\large\ { (a+b+c+d+e })^{ 2 }=\sum { { a }^{ 2 } } +\quad \sum { ab }$,

we have to prove that $\large\ 2\sum { { a }^{ 2 } } +\quad 4\sum { ab\quad \ge } \quad 5\sum { ab, }$

which is equivalent to $\large\ 2\sum { { a }^{ 2 } } \ge \quad \sum { ab }$.

The last inequality follows from $\large\ \sum { { (a-b) }^{ 2 } } \ge \quad 0.$

How is it?

. Note by Priyanshu Mishra
5 years, 2 months ago

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