prove inequality

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

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341
Hari Shankar ·

I assume u r looking for a solution:

Using Cauchy Schwarz

1+bca=a+abca2a+3abc(a+b+c)2=1+3abc1

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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