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Given $0\le a,b,c\le 1$,how to prove $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}+(1-a)(1-b)(1-c)\le 1$。Thanks。

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Check because: $\frac a {1+b+c} \leq \frac a {a+b+c}$ which will give LHS as $1 + (1-a)(1-b)(1-c)$. This contradicts the given inequality. – user62089 Mar 20 '13 at 3:22
@pondy that gives that the LHS is at most that value. It doesn't say the LHS can take that value – Thomas Andrews Mar 20 '13 at 3:57
up vote 1 down vote accepted

This is USAMO 1980 Q5, which has excellent solutions here on page 7 and here on page 2.

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thank you very much – geometryscience Mar 21 '13 at 0:58

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