Maximum of $\sum_{cyc}\frac{xy}{(x+2y)^2}$, $x,y,z>0$ and $x+y+z=3.$

Question

Let $x,y,z>0$ and $x+y+z=3$. Find the maximum of $S=\sum_{cyc}\frac{xy}{(x+2y)^2}.$

From $0\leq(\sqrt x-\sqrt{2y})^2$ we have $\frac{\sqrt{xy}}{x+2y}\leq\frac1{2\sqrt2}$ and hence $\frac{xy}{(x+2y)^2}\leq\frac18$. So we have $S\leq\frac38$. But, I suspect that $\frac13$ is the maximum of $S$ not $\frac38$.

Can you find the maximum of $S$? Thanks in advance.

Answer 1

For $x=y=z=1$ we obtain a value $\frac{1}{3}$.

We'll prove that it's a maximal value.

Indeed, we need to prove that: $$\sum_{cyc}\frac{xy}{(x+2y)^2}\leq\frac{1}{3}$$ or $$\sum_{cyc}(4x^4y^2+16x^4z^2+4x^3y^3+4x^4yz-31x^3y^2z+8x^3z^2y-5x^2y^2z^2)\geq0,$$ which is true by AM-GM.

For example, $$\sum_{cyc}(4x^4y^2+4x^2y^2z^2)\geq8\sum_{cyc}x^3y^2z,$$$$\sum_{cyc}x^4z^2\geq\sum_{cyc}x^3y^2z,$$ $$\sum_{cyc}x^4yz\geq\sum_{cyc}x^3y^2z,$$$$\sum_{cyc}x^3y^3\geq\sum_{cyc}x^3y^2z$$ and $$\sum_{cyc}x^3z^2y\geq\sum_{cyc}x^2y^2z^2.$$

Answer 2

Another way.

We need to prove that: $$\sum_{cyc}\frac{xy}{(x+2y)^2}\leq\frac{1}{3}$$ or $$\sum_{cyc}\left(\frac{xy}{(x+2y)^2}-\frac{1}{8}\right)\leq\frac{1}{3}-\frac{3}{8}$$ or $$\sum_{cyc}\frac{(2y-x)^2}{(x+2y)^2}\geq\frac{1}{3}.$$ Now, by C-S $$\sum_{cyc}\frac{(2y-x)^2}{(x+2y)^2}=\sum_{cyc}\frac{y^2(2y-x)^2}{y^2(x+2y)^2}\geq\frac{\left(\sum\limits_{cyc}(2x^2-xy)\right)^2}{\sum\limits_{cyc}y^2(x+2y)^2}.$$ Thus, it's enough to prove that: $$3\left(\sum\limits_{cyc}(2x^2-xy)\right)^2\geq\sum\limits_{cyc}y^2(x+2y)^2$$ or $$\sum_{cyc}(4x^4-6x^3y-8x^3z+13x^2y^2-3x^2yz)\geq0$$ and since $$\sum_{cyc}(3x^2y^2-3x^2yz)=\frac{3}{2}\sum_{cyc}z^2(x-y)^2\geq0,$$ it's enough to prove that: $$\sum_{cyc}2x^4-3x^3y-4x^3z+5x^2y^2)\geq0,$$ which is true by the TL method: $$\sum_{cyc}2x^4-3x^3y-4x^3z+5x^2y^2)=\sum_{cyc}x(2x^3-3x^2y+5xy^2-4y^3)=$$ $$=\sum_{cyc}\left((x-y)(2x^3-x^2y+4xy^2)-\frac{5}{4}(x^4-y^4)\right)=$$ $$=\frac{1}{4}\sum_{cyc}(x-y)^2(3x^2-6xy+5y^2)\geq0$$ and we are done.

Answer 3

Let $y=mx$, $z=\frac1n x$, $m,n>0$. It is enough to find the maximum of $$g(m,n)=\frac{m}{(2m+1)^2}+\frac{mn}{(mn+2)^2}+\frac{n}{(2n+1)^2}$$ We need to solve the system $$g_m=\frac{1-2m}{(2m+1)^3}+\frac{(2-mn)n}{(mn+2)^3}=0\tag1$$ $$g_n=\frac{1-2n}{(2n+1)^3}+\frac{(2-mn)m}{(mn+2)^3}=0\tag2$$ On the other hand, $mg_m-ng_n=0$ gives $$\frac{m(1-2m)}{(2m+1)^3}=\frac{n(1-2n)}{(2n+1)^3}.$$ From the graph of the function $h(x)=\frac{x(1-2x)}{(2x+1)^3}$, we see that given $n>0$, there is at most two $m>0$ such that $h(m)=h(n)$: $1)$ $m=n\,$ $2)$ $m\neq n.$

When $m=n$, the only positive real solution of the system is $m=n=1.$ We need to show that second case does not satisfy $(1)$ or $(2).$ The equation $(1)$ gives $$2n^2(n+4)m^4-(n^3-24n^2+16n)m^3-(24n-16)m-(2n+8)=0.$$ The constant term is $-(2n+8).$ When $n>0$ it is negative, there can not be only two positive real solutions for $m$. We are done.