work 016 – An useful result

Problem

Prove that for positive reals a,b,c

\sum \frac{a}{b+c} + \frac{4abc}{(a+b)(b+c)(c+a)} \ge 2

Proof

Clear the denominators and observe the above inequality is equivalent to –

2\sum a^3 + 6abc \ge 2 \sum a^2(b+c)

which is the schur’s inequality for degree 1 .

Hence proved with equality for a=b=c or a=b , c=0

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