Prove (preferably by induction) that $$x^{\frac{n(n-1)}{2}}y^{\frac{n(n-1)}{2}}(x^{n}+y^{n})\leq2$$ where $x,y>0;x+y=2$ and $n$ is a positive integer.
Some specific cases of this are: $$xy(x^2+y^2)\leq2,$$ $$x^{3}y^{3}(x^{3}+y^{3})\leq2,$$ $$x^{6}y^{6}(x^{4}+y^{4})\leq2.$$
They all can be proven pretty easily by AM-GM inequality.
I have also noticed that there's a high chance $x^{\frac{n(n-1)}{2}}y^{\frac{n(n-1)}{2}}(x^{n}+y^{n})\geq x^{\frac{(n+1)n}{2}}y^{\frac{(n+1)n}{2}}(x^{n+1}+y^{n+1})$ but haven't come up with a proof yet.
We need to prove that: $$x^{\frac{n(n-1)}{2}}y^{\frac{n(n-1)}{2}}(x^n+y^n)\leq2\left(\frac{x+y}{2}\right)^{n^2}.$$ Since the last inequality is homogeneous and symmetric, we can assume that $y=1$ and $x\geq1$
and we need to prove that $f(x)\geq0,$ where $$f(x)=n^2\ln(x+1)-\ln(x^n+1)-\frac{n(n-1)}{2}\ln x-(n^2-1)\ln2.$$ Now, $$f'(x)=\frac{n((n-1)x^{n+1}-(n+1)x^n+(n+1)x-n+1)}{2x(x+1)(x^n+1)}.$$ Let $g(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-n+1.$
Thus, by AM-GM $$g'(x)=(n+1)((n-1)x^n-nx^{n-1}+1)\geq(n+1)\left(n\sqrt[n]{(x^n)^{n-1}\cdot1}-nx^{n-1}\right)=0.$$ Id est, $$g(x)\geq g(1)=0,$$ $$f(x)\geq f(1)=0$$ and we are done!