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
 A: 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!
A: I changed the letters $a,b,c$ to $x,y,z$ without any good reason.
Let $S = x+y+z, S_2 = xy + yz + zx, P = xyz$.
As the problem is invariant under different permutations of $x,y,z$, then you can express everything with these mew variables. Indeed:
$$
\frac 1{2+x}+\frac 1{2+y}+\frac 1{2+z} = \frac{S_2 + 4S + 12}{P+2S_2 + 4S + 8}
\\
x+y+z = \frac 1x+\frac 1y+\frac 1z \iff SP = S_2
$$
Now you are left with a two variables function to study.
A: the given inequality is equivalent to
$abc+ab+ac+bc\geq 4$ (I)from the given condition we get
$abc(a+b+c)=ab+ac+bc$ (1)
this gives in our inequality (I)
$abc+abc(a+b+c)\geq 4$
since $a+b+c\geq 3$ we have in the case $abc\geq 1$:
$abc+abc(a+b+c)\geq 1+3\cdot 1=4$
now our second case:
$abc\le 1$
then we can set $a=\frac{1}{a'},b=\frac{1}{b'},c=\frac{1}{c'}$
and our condition $a'+b'+c'=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}$ is fulfilled and we can make the same like in part I.
