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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?

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Note by Priyanshu Mishra
1 year, 11 months ago