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
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It is weird that i am answering my question but finally i got the solution $$(a+b+c)\left(\frac {1}{1+ab}+\frac {1}{1+bc}+\frac {1}{1+ca}\right)$$
$$=\frac {a}{1+bc}+\frac {b}{1+ca}+\frac {c}{1+ab}+\frac {a+b}{1+ab}+\frac {b+c}{1+bc}+\frac {c+a}{1+ca} \leq\frac {a}{1+bc}+\frac {b}{1+ca}+\frac {c}{1+ab}+3$$
$$\leq\frac{a}{1+bc}-\frac{ca}{b+ca}-\frac{ab}{c+ab}+5 \leq a\left(1-\frac{c}{b+c}-\frac{b}{c+b}\right)+5 = 5.$$
Your way is wrong. Try $b=c\rightarrow0^+$.
The following reasoning gives a proof.
We need to prove that $$\sum_{cyc}\frac{a+b+c}{1+ab}\leq5$$ or $$\sum_{cyc}\frac{c}{1+ab}\leq2+\sum_{cyc}\left(1-\frac{a+b}{1+ab}\right)$$ or $$\sum_{cyc}\frac{c}{1+ab}\leq2+\sum_{cyc}\frac{(1-a)(1-b)}{1+ab}$$ and it's enough to prove that: $$\sum_{cyc}\frac{c}{1+ab}\leq2.$$ Now, let $f(a,b,c)=\sum\limits_{cyc}\frac{c}{1+ab}.$
Thus, $f$ is a convex function of $a$, of $b$ and of $c$.
Id est, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,0)=2$$ and we are done!
For example, $$\frac{\partial^2f}{\partial a^2}=\frac{2cb^2}{(1+ab)^3}+\frac{2bc^2}{(1+ac)^3}\geq0.$$