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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.