Mathematical Reflections 2014, Issue 1 - Problem J292

Problem:
Find the least real number $k$ such that for every positive real numbers $x, y, z$, the following
inequality holds: $$\prod_{cyc} (2xy+yz+zx) \leq k(x+y+z)^6.$$

Proposed by Dorin Andrica.

Solution:
Observe that $$(x-y)^2+z^2 \geq 0 \iff x^2+y^2+z^2-2xy \geq 0 \iff (x+y+z)^2 \geq 4xy+2yz+2zx,$$ so $2xy+yz+zx \leq \dfrac{1}{2}(x+y+z)^2$. Likewise, $2yz+zx+xy \leq \dfrac{1}{2}(x+y+z)^2$ and $2zx+xy+yz \leq \dfrac{1}{2}(x+y+z)^2$. Therefore,
$$\prod_{cyc} (2xy+yz+zx) \leq \dfrac{1}{8}(x+y+z)^6.$$ The equality is attained if and only if $(x-y)^2+z^2=0, (y-z)^2+x^2=0, (z-x)^2+y^2=0$, i.e. if and only if $x=y=z=0$. Clearly, $k_{\min}=1/8$, because for each inequality $1/2$ was the least real number for which the inequality was satisfied.


Note. Actually, this is not the minimum value of $k$, which is $64/729$. See the official solution.