INEQUALITY'S BLOG

July 16, 2011

Problem 238 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 11:13 am

Prove that for a, b, c>0 we have:

\displaystyle \frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\displaystyle \geq \frac{9}{4}\biggl(\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\biggl)

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