This following problem is from my Conjecture many years ago,

Question : Let $a,b>0,n\in N^{+},n\ge 3$,such $$a^n+b^n+(2n+2)(ab)^n\le 2n$$

Conjecture: then $a+b\le 2$ or

$a+b>2.a>0.b>0,n\ge 3$,then we prove $$a^n+b^n+(2n+2)a^nb^n-2n>0$$ or

it suffuce to prove $$a^n+b^n+(2n+2)a^nb^n-2n>0\rm{when}~ a+b=2$$

This inequality is from this conjecture $$\sqrt[n]{\dfrac{x}{1+y}}+\sqrt[n]{\dfrac{y}{1+x}}\le 2,\rm{where}~ x+y\le 2n,n\ge 3,n\in N^{+}$$ it is not hard to prove this case $n=3,4$,but for $n\ge 5$, use this $$\sqrt[n]{\dfrac{x}{1+y}}=a,\sqrt[n]{\dfrac{y}{1+x}}=b$$ then $$x=\dfrac{a^n(b^n+1)}{1-(ab)^n},y=\dfrac{b^n(a^n+1)}{1-(ab)^n}$$take $x+y\le 2n\Longrightarrow a^n+b^n+(2n+2)(ab)^n\le 2n$