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

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.
