# Prove this conjecture inequality

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 $x+y\le 2n\Longrightarrow a^n+b^n+(2n+2)(ab)^n\le 2n$

Take $a=1+x,b=1-x$, $|x|\leqslant 1$. We have to prove $(1+x)^n+(1-x)^n+(2n+2)(1-x^2)^n\geqslant 2n$. By Bernoulli inequality we have $(1-x^2)^n\geqslant 1-nx^2$ and by binomial expansion $(1+x)^n+(1-x)^n=2+n(n-1)x^2+2\binom{n}4 x^4+\dots$ So, it suffices to prove that $$2+n(n-1)x^2+2\binom{n}4 x^4+2n+2-n(2n+2)x^2\geqslant 2n,$$ this is quadratic in $x^2$, discriminant equals $(n^2+3n)^2-32\binom{n}4$, this is negative for large enough $n$ (about $n\geqslant 42$, I think). Well, this is not quite satisfactory, but at least it works for all but finitely many $n$. Next trick is generalized Bernoulli $(1-x^2)^n\geqslant 1-nx^2+\binom{n}2 x^4-\binom{n}3x^6$ (this follows from your favourite remainder term in Taylor's expansion for $(1-t)^n$). Assume that $n$ satisfies $2\binom{n}6\geqslant (2n+2)\binom{n}3$, this is true for $n\geqslant 17$. Then it suffices to prove $$4-(n^2+3n)x^2+\left(2\binom{n}4+(2n+2)\binom{n}2\right)x^4\geqslant 0.$$ For $n\geqslant 17$ this is true, so we have already proved your inequality for $n\geqslant 17$. Well, for $n=14,15,16$ the term with $x^8$ allows to beat negative term with $x^6$ (that is, after we subtract $(2-\frac{n^2+3n}4x)^2$ from our expression and divide by $x^4$, we get a positive quadratic trinomial in $x^2$.) Computer plots confirm that for $3\leqslant n\leqslant 13$ conjecture also holds, but this is probably not what you need.