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

• Clean up the denominators. Commented Oct 31, 2014 at 21:34

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!

• But that combination does not satisfy the constraint. When $c=0.01$ if $a=b$ then they are both $a=b=50.015$ and the cyclic sum is about $0.536$. Commented May 26, 2017 at 16:10
• @Mark Fischler I fixed my post. See now, please. Commented May 26, 2017 at 18:02
• You got it! Nice idea, about exploring the situation when the inequality itself is tight. Commented May 31, 2017 at 15:03
• @Mark Fischler Thank you for your interest. Commented May 31, 2017 at 15:43

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.

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.

• I think, making the same like in part I, we get $$a'b'c'+a'b'c'(a'+b'+c')\ge 4$$ not $$abc+abc(a+b+c)\ge 4$$ Commented Nov 1, 2014 at 8:29
• yes of course it is the same Commented Nov 1, 2014 at 11:13
• But proving the inequality for $a\{a',b',c'\}$ does not prove the inequality for $\{a,b,c\}$. Commented May 26, 2017 at 16:04