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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?

Please do reply and post solutions if any.

.

Note by Priyanshu Mishra
2 years, 2 months ago

No vote yet
1 vote

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