prove inequality

if \ a+b+c =1 ,; a,b, c \rightarrow R^+ \\ prove \ that \ \\ \\ \sum_{cyclic}{} {\frac{a}{1+bc}}\geq \frac{9}{10}

1 Answers

341

Hari Shankar ·2010-01-13 08:43:24

I assume u r looking for a solution:

Using Cauchy Schwarz

\sum \frac{a}{1+bc} = \sum \frac{a^2}{a+abc} \ge \frac{(a+b+c)^2}{\sum a + 3abc} = \frac{1}{1+3abc}

From AM-GM a+b+c =1 \Rightarrow 3abc \le \frac{1}{9} \Rightarrow \frac{1}{1+3abc} \ge \frac{9}{10}

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