I have found that this is true for $p=1, 2$ and was trying to show the function $f(p) = (a+b)^p + (a-b)^p -2a^p - p(p-1)a^{p-2}b^2 \geq 0$ for $p \in (1,2) $ but have been unable to do so. I differentiated it twice to show that concavity on the interval would guarantee so but was unable to prove so.
Also I know $(a+b)^p + (a-b)^p \geq 2a^p$ by convexity of $x^p$ but haven't found a way to account for the last term. It looks a lot like a derivative but I'm not quite able to proceed with that knowledge.
$f(p) = (a+b)^p + (a-b)^p -2a^p - p(p-1)a^{p-2}b^2 \geq 0 $
Letting $g(p) = f(p)/a^p $ and $c = b/a$, $g(p) = (1+c)^p + (1-c)^p -2 - p(p-1)c^2 $.
Let $h(c) = (1+c)^p + (1-c)^p -2 - p(p-1)c^2 $. $h(0) = 0$.
$h'(c) =p(1+c)^{p-1}-p(1-c)^{p-1}-2p(p-1)c =p((1+c)^{p-1}-(1-c)^{p-1}-2(p-1)c) $ so $h'(0) = 0$.
Then $h''(c) =p(p-1)((1+c)^{p-2}+(1-c)^{p-2}-2) $.
$h''(0) = 0$ and $h'''(c) =p(p-1)(p-2)((1+c)^{p-3}-(1-c)^{p-3}) \gt 0 $ since $p(p-1)(p-2) < 0$, $0 < (1+c)^{p-3} < 1$, and $(1-c)^{p-3} > 1$.
Therefore, working our way back, $h''(c) > 0$ for $0 < c < 1$, $h'(c) > 0$ for $0 < c < 1$, and, finally, $h(c) > 0$ for $0 < c < 1$.