How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$? How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$
for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$.
I do not know where to start. I need some idea and advice on this problem.Thanks

Let $\sum\limits_{cyc}\frac{1}{2+a}>1$ and $a=ka'$ such that $k>0$ and
$$\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}=1.$$
Hence,
$$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}>1=\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}$$ or
$$\frac{1}{2+ka'}>\frac{1}{2+a'},$$
which gives $k<1$.
In another hand,
$$a'+b+c-\frac{1}{a'}-\frac{1}{b}-\frac{1}{c}=\frac{a}{k}+b+c-\frac{1}{\frac{a}{k}}-\frac{1}{b}-\frac{1}{c}>a+b+c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=0,$$
which is contradiction because we'll prove now that
$$a'+b+c-\frac{1}{a'}-\frac{1}{b}-\frac{1}{c}\leq0.$$
Indeed, let $a'=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ are positives.
Hence, the condition $\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}=1$ gives $c=\frac{2z}{x+y}$ and we need to prove that
$$\sum_{cyc}\frac{y+z}{2x}\geq\sum_{cyc}\frac{2x}{y+z}$$ or
$$\sum_{cyc}x\left(\frac{1}{y}+\frac{1}{z}\right)\geq\sum_{cyc}\frac{4x}{y+z},$$
which is C-S:
$$\frac{1}{y}+\frac{1}{z}\geq\frac{(1+1)^2}{y+z}=\frac{4}{y+z}.$$
Done!
