# Symmetric Inequality involving square roots of two variables

Let $x$ and $y$ be positive numbers. Find the the minimum value of the following expression. $$\left(x^2+xy +2y^2 \right)^\frac{1}{2} +\left(2x^2+xy +y^2 \right)^\frac{1}{2} +\left(3x^2+xy +5y^2 \right)^\frac{1}{2} +\left(5x^2+xy +3y^2 \right)^\frac{1}{2}$$ Subject to: $$x+y = 2016$$

I can see that the inequality is symmetric in $x$ and $y$, and the minimum seems to attained, when $x=y=1008,$ but I don't know, how to show this.

Any help is appreciated.

• If it is symmetric, the extrema can only be where $x=y$. If it is not a maximum, it is a minimum. – N74 Jul 29 '17 at 10:56

I set $a=1008$ because of the symmetry, and then $x=a+t$ and $y=a-t$. I minimize just the first two terms of the expression, which I set equal to $f$. Then $f$ is minimized when $f^2$ is. If I plug the above expressions in $f^2$, and simplify, I get
$$f^2 = 4t^2+8a^2+4(t^4+3t^2+4a^4)^{1/2}$$
which obviously has its least value when $t=0$. Then I set $g$ equal to the other two terms and do the same thing and discover that $g$ is also minimized when $t=0$. I conclude that $x=a$ and $y=a$.
By Minkowski $$\sqrt{x^2+xy+2y^2}+\sqrt{2x^2+xy+y^2}+\sqrt{3x^2+xy+5y^2}+\sqrt{5x^2+xy+3y^2}=$$ $$=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{7}{4}y^2}+\sqrt{\left(y+\frac{x}{2}\right)^2+\frac{7}{4}x^2}+$$ $$+\sqrt{3\left(x+\frac{y}{6}\right)^2+\frac{59}{12}y^2}+\sqrt{3\left(y+\frac{x}{6}\right)^2+\frac{59}{12}x^2}\geq$$ $$\geq\sqrt{\left(\frac{3(x+y)}{2}\right)^2+\frac{7}{4}(x+y)^2}+\sqrt{3\left(\frac{7(x+y)}{6}\right)^2+\frac{59}{12}(x+y)^2}=10080.$$ The equality occurs for $x=y=1008$, which says that $10080$ is a minimal value.