# Prove that $\sum_{cyc}\frac{1}{(2a+2b+c)^3}\le\frac{3}{125x}$ if $abc=x$

How can you prove that for positive reals a, b, c, $\sum_{cyc}\frac{1}{(2a+2b+c)^3}\le\frac{3}{125x}$ given that $abc=x$?

I attempted AM-GM and failed, but it seems pretty clear that maximum value is reached when $a=b=c$. I'm having some trouble proving this.

edit: To expand on this, how to show that $\sum_{cyc}\frac{1}{(na+nb+c)^3}\le\frac{3}{(2n+1)^3x}$?

credits to/inspired by: The maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$

• Note that 125=5^3 and 2+2+1=5.You may try weighted AM.GM.inequality. – Beatty1729 Oct 29 '17 at 7:15
• After some calculation I get this: $$28([5,4,0]+[4,3,2]+[5,3,1])\leq 13[3,3,3]+47[4,4,1]+24[6,3,0]$$ but I have no idea how to prove this. Probably some Schur inequality must be involved, but I'm not sure. – Aqua Oct 29 '17 at 11:23

By C-S and AM-GM we obtain \begin{align} &\sum_{cyc}\frac{1}{(2a+2b+c)^3} \\ =&\sum_{cyc}\frac{1}{8a^3+8b^3+c^3+24a^2b+24ab^2+12a^2c+12b^2c+6c^2a+6c^2b+24abc} \\ =&\sum_{cyc}\frac{1}{8a^3+8b^3+c^3+24ab(a+b+c)+6c(2(a^2+b^2)+(a+b)c)} \\ \leq&\sum_{cyc}\frac{1}{8a^3+8b^3+c^3+24ab(a+b+c)+6c((a+b)^2+(a+b)c)} \\ \leq&\frac{1}{(17+72+36)^2}\sum_{cyc}\left(\frac{17^2}{8a^3+8b^3+c^3}+\frac{72^2}{24ab(a+b+c)}+\frac{36^2}{6c(a+b)(a+b+c)}\right) \\ =&\frac{289}{125^2}\sum_{cyc}\frac{1}{8a^3+8b^3+c^3}+\frac{216}{125^2abc}+\frac{216}{125^2(a+b+c)abc}\sum_{cyc}\frac{ab}{a+b}\\ \leq&\frac{289}{125^2}\sum_{cyc}\frac{1}{8a^3+8b^3+c^3}+\frac{216}{125^2abc}+\frac{216}{125^2(a+b+c)abc}\sum_{cyc}\frac{\frac{(a+b)^2}{4}}{a+b} \\ =&\frac{289}{125^2}\sum_{cyc}\frac{1}{8a^3+8b^3+c^3}+\frac{324}{125^2abc}. \end{align}

Thus, it remains to prove that $$\frac{289}{125^2}\sum_{cyc}\frac{1}{8a^3+8b^3+c^3}+\frac{324}{125^2abc}\leq\frac{3}{125abc}$$ or $$\sum_{cyc}\frac{1}{8a^3+8b^3+c^3}\leq\frac{3}{17abc}$$ or $$\sum_{cyc}\frac{1}{8a+8b+c}\leq\frac{3}{17},$$ where $a$, $b$ abd $c$ are positives such that $abc=1$.

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

Thus, we need to prove the third degree symmetric inequality,

which says that we need to prove that $f(v^2)\geq0$, where $f$ is a linear function.

But the linear function gets a minimal value for an extremal value of $v^2$,

which happens for equality case of two variables.

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

We obtain $$(a-1)^2(432a^4+167a^3-98a^2+48a+24)\geq0,$$ which is obvious.

Done!

Another way.

Let $abc=1$.

Thus, by AM-GM $$\sum_{cyc}\frac{1}{(2a+2b+c)^3}=\sum_{cyc}\frac{1}{(3\cdot\left(\frac{a+b+c}{3}\right)+a+b)^3}\leq\sum_{cyc}\frac{1}{125\sqrt[5]{\left(\frac{a+b+c}{3}\right)^9a^3b^3}}=$$ $$=\sum_{cyc}\frac{3}{125(a+b+c)\sqrt[5]{\left(\frac{a+b+c}{3}\right)^4a^3b^3}}\leq\frac3{125}\sum_{cyc}\frac1{(a+b+c)\sqrt[5]{a^3b^3}}.$$ Thus, it's enough to prove that $$a+b+c\geq\sum_{cyc}\frac{1}{\sqrt[5]{a^3b^3}}$$ or $$a+b+c\geq\sum_{cyc}\sqrt[5]{\frac{a^3b^3c^3}{a^3b^3}}$$ or $$a+b+c\geq\sum_{cyc}a^{\frac{3}{5}}$$ or $$a+b+c\geq\sum_{cyc}a^{\frac{11}{15}}b^{\frac{2}{15}}c^{\frac{2}{15}},$$ which is true by Muirhead because $(1,0,0)\succ\left(\frac{11}{15},\frac{2}{15},\frac{2}{15}\right).$

• Thanks. Would it be possible to prove a more generalized version of this problem, $\sum_{cyc}\frac{1}{(na+nb+c)^3}\le\frac{3}{(2n+1)^3x}$ – long_live Oct 29 '17 at 17:47
• Sorry...I'm new to the site. I clicked the check mark? – long_live Oct 29 '17 at 18:25
• It's OK. I think your new inequality is wrong. Open please another topic with your new problem. – Michael Rozenberg Oct 29 '17 at 18:27
• Would you please define $f$ and explain the rationale for the sentence "...which happens for equality case of two variables"? – Hans Feb 25 '18 at 4:12
• It's $\sum\limits_{cyc}\frac{1}{8(a+b+c)-7a}\leq\frac{3}{17w}$ or $\sum\limits_{cyc}\frac{1}{24u)-7a}\leq\frac{3}{17w}$ or $1728u^3+3528uv^2-343w^3\geq17w(240u^2+49v^2)$, which gives $f(v^2)=1728u^3+3528uv^2-343w^3-17w(240u^2+49v^2)$, which is a linear function. The fact that it's a linear function we can see without full expanding. – Michael Rozenberg Feb 25 '18 at 6:04