Weighted AM-GM: $(1)$ $\frac{3}{4}x^4+\frac{1}{4}y^4\geq x^3y$ $(2)$ $\frac{1}{4}x^4+\frac{3}{4}y^4\geq xy^3$ $x,y>\sqrt{2} \rightarrow x^2,y^2>2$ $(3)$ $x^2y^2=\frac{1}{2}x^2y^2+\frac{1}{2}x^2y^2>\frac{1}{2}x^2\cdot 2+\frac{1}{2}\cdot 2 \cdot y^2=x^2+y^2$ Adding $(1)+(2)+(3)$ gives the desired inequality.

Another variant: The left side is equal $\frac{x^5+y^5}{x+y}$.Given the condition we have $x^5+y^5>2(x^3+y^3)$ and $\frac{x^5+y^5}{x+y}>\frac{2(x^3+y^3)}{x+y}=2(x^2-xy+y^2){\ge}x^2+y^2$.Ok

Pirkuliyev Rovsen wrote: Another variant: The left side is equal $\frac{x^5+y^5}{x+y}$.Given the condition we have $x^5+y^5>2(x^3+y^3)$ and $\frac{x^5+y^5}{x+y}>\frac{2(x^3+y^3)}{x+y}=2(x^2-xy+y^2){\ge}x^2+y^2$.Ok did you use that $x> \sqrt 2$ and $y> \sqrt 2$ in your proof ??

Yes, look at the first inequality

a lot thanks!