The third degree Schur helped! \begin{align*} 2(\text{left}-\text{right})\prod(x^2+y^2)=3\prod(x-y)^2+2\sum xy(x-y)^4\\ +32x^2y^2z^2+8xyz\sum x(x-y)(x-z)+8xyz\sum x(y-z)^2\ge0. \end{align*}

oVlad wrote: Let $x,y,z{}$ be positive real numbers. Prove that \[(xy+yz+zx)\left(\frac{1}{x^2+y^2}+\frac{1}{y^2+z^2}+\frac{1}{z^2+x^2}\right)>\frac{5}{2}.\] Moldova TST 2013 here With $a,b,c\ge 0$ prove that $$(ab+bc+ca)(\frac{1}{a^2+b^2}+\frac{1}{c^2+b^2}+\frac{1}{a^2+c^2})\ge \frac{5}{2}+\frac{16a^2b^2c^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)} $$

sqing wrote: With $a,b,c\ge 0$ prove that $$(ab+bc+ca)(\frac{1}{a^2+b^2}+\frac{1}{c^2+b^2}+\frac{1}{a^2+c^2})\ge \frac{5}{2}+\frac{16a^2b^2c^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)} $$ To prove this, use my SOS, remove the "$32x^2y^2z^2$" will do!

Assume that $x\ge y\ge z$. Then $\ \ \frac{xy+xz}{x^2+z^2}\ge \frac{y}{x}, \ \ \ \frac{xy+xz}{y^2+z^2}\ge \frac{x}{y}$. $(xy+yz+zx)\left(\frac{1}{x^2+y^2}+\frac{1}{y^2+z^2}+\frac{1}{z^2+x^2}\right) > $ $>\frac{xy}{x^2+y^2}+\frac{xy+xz}{x^2+z^2}+\frac{xy+xz}{y^2+z^2}\ge \frac{xy}{x^2+y^2}+\frac{y}{x}+\frac{x}{y}=$ $= \frac{xy}{x^2+y^2}+\frac{x^2+y^2}{xy}=\frac{1}{a}+a=\frac{(a-2)(2a-1)}{2a}+\frac{5}{2}\ge \frac{5}{2}$ where $\ \ a=\frac{x^2+y^2}{xy}\ge 2$. PS: If I am not wrong, this solution belongs to quykhtn-qa1 .