Arqady post this before.

Prove or disprove $$\frac{1}{a+bc+4}+\frac{1}{b+ca+4}+\frac{1}{c+ab+4}\le \frac{1}{2}$$holds for all positive reals $a,b,c$ such that $a^2+b^2+c^2+abc=4$.

sqing wrote: Prove or disprove $$\frac{1}{a+bc+4}+\frac{1}{b+ca+4}+\frac{1}{c+ab+4}\le \frac{1}{2}$$holds for all positive reals $a,b,c$ such that $a^2+b^2+c^2+abc=4$. It's wrong. Try $a=2$ and $b=c=0$.

sqing wrote: Prove or disprove $$\frac{1}{a+bc+4}+\frac{1}{b+ca+4}+\frac{1}{c+ab+4}\le \frac{1}{2}$$holds for all positive reals $a,b,c$ such that $a^2+b^2+c^2+abc=4$. Reverse inequality is true.

erzhane wrote: sqing wrote: Prove or disprove $$\frac{1}{a+bc+4}+\frac{1}{b+ca+4}+\frac{1}{c+ab+4}\le \frac{1}{2}$$holds for all positive reals $a,b,c$ such that $a^2+b^2+c^2+abc=4$. Reverse inequality is true. Yes, because $ab+ac+bc\leq2+abc$ and $a+b+c\leq3.$

Thanks. Then $$2\le \frac{4}{a+bc+4}+\frac{4}{b+ca+4}+\frac{4}{c+ab+4}\le 1+\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$$holds for all positive reals $a,b,c$ such that $a^2+b^2+c^2+abc=4.$

Hmm, any proof?

$\frac 4{a+bc+4}\le \frac 1{2a+1}+\frac a{bc+2a}.\Box$