# Show that $\sum_{cyc}\frac{1}{(na+nb+c)^3}\le\frac{3}{(2n+1)^3x}$?

How can you prove that for positive reals $a, b, c$ with $abc=x$, $\sum_{cyc}\frac{1}{(na+nb+c)^3}\le\frac{3}{(2n+1)^3x}$ (if this is even true)? It's been proved for $n=2$.

It seems like equality holds when $a=b=c$, but I'm not sure how to show this. Attempted AM-GM but it didn't work out nicely.

This inequality is wrong for all natural $n\geq8$.

Try $a=b=0.547$ and $c=\frac{1}{a^2}.$

A proof for $n=7$.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$\sum_{cyc}\frac{1}{(7a+7b+c)^3}\leq\frac{1}{1125abc}$$ or $$\frac{64w^6+6776u^3w^3+96uv^2w^3+39445u^6-32340u^4v^2+1008u^2v^4+576v^6}{(49u^3+84uv^2-8w^3)^3}\leq\frac{1}{125w^3}$$ or $f(w^3)\geq0,$ where $$f(w^3)=(49u^3+84uv^2-8w^3)^3-$$ $$-125w^3(64w^6+6776u^3w^3+96uv^2w^3+39445u^6-32340u^4v^2+1008u^2v^4+576v^6).$$ But, $$f'(w^3)=-24(49u^3+84uv^2-8w^3)^2-$$ $$-125(64w^6+6776u^3w^3+96uv^2w^3+39445u^6-32340u^4v^2+1008u^2v^4+576v^6)-$$ $$-125w^3(128w^3+6776u^3+96uv^2)<0,$$ which says that $f$ decreases.

It est, it's enough to prove our inequality for a maximal value of $w^3$,

which happens for an equality case of two variables.

Since our inequality is homogeneous and symmetric we can assume that $b=c=1$, which gives $$(a-1)^2(343a^4+16268a^3-103512a^2+48848a+1404928)\geq0,$$ which is obvious enough.

Done!

• Oh, I see. I tried it for some lower numbers and assumed it worked...guess not. Is it possible to prove it for specific values such as $n=4$, with $abc=1$ (sorry), thanks a lot for your help! – long_live Oct 29 '17 at 19:14
• @long_live I think we can prove this inequality by the $uvw$ method even for $n=7$ but it's very ugly. – Michael Rozenberg Oct 29 '17 at 19:15
• I'll try that method, thanks! – long_live Oct 29 '17 at 19:17
• @long_live For $abc=1$ it's the same. It seems even for $n=3$ it's very ugly. – Michael Rozenberg Oct 29 '17 at 19:18