Is it true that $ (abc+xyz)^{\frac13} \le \frac13 \left( (ab+xy)^\frac12 + (bc+yz)^\frac12 +(ac+xz)^\frac12 \right)$

Question

Given $6$ positive real numbers $a,b,c,x,y,z>0$ show that $$ (abc+xyz)^{\frac13} \le \frac13 \left( (ab+xy)^\frac12 + (bc+yz)^\frac12 +(ac+xz)^\frac12 \right)$$ For context, this is a specific case of a more general conjectured inequality, and I hoped that solving a simpler example would help for the full solution.

I wasn't able to find a single counterexample, but I may just have missed it. All my attempts at proving this started with the inequality \begin{equation} \label{eq} (abc+xyz)^{\frac13} \le (abc)^\frac13 +(xyz)^\frac13 \tag{*} \end{equation} since the terms on the right are easier to deal with, as one can use $$(abc)^\frac13 \le \frac{a+b+c}{3}$$ or the equivalent $$ (abc)^\frac13 \le (ab)^\frac12 + (bc)^\frac12 +( ac)^\frac12$$ The problem is, in (\ref{eq}), the right hand side has already exceeded the goal. One can easily show that it can be bigger than the r.h.s in the original inequality.

In the end, I am stuck. I am not skilled in inequalities, so I ask for help. Thank you in advance.

Answer 1

That inequality does hold and is in fact strict for $a, b, c, x, y, z > 0$.

It suffices to prove that $$ \tag{$*$} (abc+xyz)^2 < (ab+xy)(bc+yz)(ca+zx) \, . $$ Then $$ \begin{align} (abc+xyz)^{1/3} &< \left( (ab+xy)^{1/2}(bc+yz)^{1/2}(ca+zx)^{1/2}\right)^{1/3} \\ &\le \frac 13 \left( (ab+xy)^{1/2}+(bc+yz)^{1/2}+(ca+zx)^{1/2}\right) \end{align} $$ by the inequality between the geometric and the arithmetic mean.

And $(*)$ is true because $$ \begin{align} (abc+xyz)^2 &= (abc)^2 + 2abcxyz + (xyz)^2 \\ &\le (abc)^2 + abxy(c^2+z^2) + (xyz)^2 \\ &= ab \cdot bc \cdot ca + xy \cdot bc \cdot ca + ab \cdot yz \cdot zx + xy \cdot yz \cdot zx \end{align} $$ and the expansion of the right-hand side of $(*)$ contains all those terms, plus more strictly positive terms.