TITLE: Let $a$, $b$, $c$ be three positive real numbers contained in the interval [0,1].
QUESTION [0 upvotes]: Prove that $$\frac {1}{1+ab}+\frac {1}{1+bc}+\frac {1}{1+ca}\leq \frac {5}{a+b+c}$$
Now I have proceed thus that $$(a+b+c)\left (\frac {1}{1+ab}+\frac {1}{1+bc}+\frac {1}{1+ca}\right)\leq (a+b+c)\left (\frac {1}{ab+c}+\frac {1}{bc+a}+\frac {1}{ca+b} \right )$$

I have this part of solution.I want to find this solution...If you understand this please tell me

REPLY [0 votes]: It is weird that i am answering my question but finally i got the solution
$$(a+b+c)\left(\frac {1}{1+ab}+\frac {1}{1+bc}+\frac {1}{1+ca}\right)$$
$$=\frac {a}{1+bc}+\frac {b}{1+ca}+\frac {c}{1+ab}+\frac {a+b}{1+ab}+\frac {b+c}{1+bc}+\frac {c+a}{1+ca} \leq\frac {a}{1+bc}+\frac {b}{1+ca}+\frac {c}{1+ab}+3$$
$$\leq\frac{a}{1+bc}-\frac{ca}{b+ca}-\frac{ab}{c+ab}+5 \leq a\left(1-\frac{c}{b+c}-\frac{b}{c+b}\right)+5 = 5.$$
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