If $a+b+c = 3abc$ and $\frac17 \leq k \leq 7$ prove $ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $

@Michael Rozenberg, in If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$, asks for a proof of one special case ($k=7$) of what I believe is a more general set of identities:

For positive $(a,b,c)$ with $a+b+c = 3abc$, and $\frac17 \leq k \leq 7$, $$ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $$

Some hairy calculation which I would hardly call a proof hinted to me that this inequality holds for all $\frac17 \leq k \leq 7$. A bit of numerical work indicates that it holds for $k\geq0.1366$; it does not hold for $k \leq 0.1365$.

Similarly, the inequality holds for $k\leq7.3242735840783$; it does not hold for $k \geq 7.3242735840784$.

For example, for $k=\frac18$, we have as a counterexample $(a=4,b=\frac54,c=\frac38)$, for which $$ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} = \frac{11892}{4403}>\frac83=\frac3{k+1} $$ A counterexample for $k=8$ has $(a=2,b=\frac12,c=\frac54)$, for which $$ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} = \frac{103}{308}>\frac13=\frac3{k+1} $$

My question is, prove that highlighted inequality.

Although tagged as "contest math" I have not actually seen this problem in a contest.

Solution 1:

In fact it's very simple we have an equivalent like this :

Let $a,b,c>0$ then : $$\sqrt{\frac{abc}{a+b+c}}\Big(\sum_{cyc}\frac{1}{7a+b}\Big)\leq\frac{\sqrt{3}}{8}$$

As usual the inequality of Michael Rozenberg is optimized so we have for $1\leq k\leq 7$:

Let $a,b,c>0$ then we have : $$\sqrt{\frac{abc}{a+b+c}}\Big(\sum_{cyc}\frac{1}{ka+b}\Big)\leq\sqrt{\frac{abc}{a+b+c}}\Big(\sum_{cyc}\frac{1}{7a+b}\Big)\frac{8}{k+1}\leq\frac{\sqrt{3}}{8}\frac{8}{k+1}$$

It's not so hard to prove and for the case of $\frac{1}{7}\leq k\leq 1$ I let you guess the right substitution wich take back to the case where $1\leq k \leq 7$ .

Remains to prove the first inequality of my answer wich have been done by brute force in the post of Michael Rozenberg .