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

 ab+c+bc+d+cd+e+de+a+ea+b≥52\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:

 ab+c+bc+d+cd+e+de+a+ea+b=a2ab+ac+b2bc+bd+c2cd+ce+d2de+ad+e2ae+be≥(a+b+c+d+e)2∑ab\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  (a+b+c+d+e)2=∑a2+∑ab\large\ { (a+b+c+d+e })^{ 2 }=\sum { { a }^{ 2 } } +\quad \sum { ab },

we have to prove that  2∑a2+4∑ab≥5∑ab,\large\ 2\sum { { a }^{ 2 } } +\quad 4\sum { ab\quad \ge } \quad 5\sum { ab, }

which is equivalent to  2∑a2≥∑ab\large\ 2\sum { { a }^{ 2 } } \ge \quad \sum { ab }.

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

How is it?

Please do reply and post solutions if any.

.

Note by Priyanshu Mishra
5 years, 8 months ago

No vote yet
1 vote

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