<s>[INST] How to prove $\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}$ For real numbers $a, b$ and all $1\le p\le 2$, how to prove $$\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}?$$
 [/INST] let $m=|a+b|,n=|a-b|$
first we prove $\left (\dfrac{m^p+n^p}{2} \right)^2 \ge \left(p\left(\dfrac{m-n}{2}\right)^2+mn\right)^p$
WLOG, let $ m\ge n , x=\dfrac{m}{n}\ge 1  \iff  \left (\dfrac{x^p+1}{2} \right)^2 \ge \left(p\left(\dfrac{x-1}{2}\right)^2+x\right)^p \iff f(x)=2\ln{\left (\dfrac{x^p+1}{2} \right)}-p\ln{\left ( \left( \dfrac{x-1}{2}\right)^2+x \right)} \ge 0$
$f'(x)=\dfrac{2px((2-p)x^p+px^{p-1}-px-2+p)}{x^p(px^3+(4-2p)x^2+px)+px^3+(4-2p)x^2+px} $
$g(x)=(2-p)x^p+px^{p-1}-px-2+p, \\g'(x)=p(2-p)x^{p-1}+p(p-1)x^{p-2}-p,\\g''(x)=p(2-p)(p-1)(x-1)x^{p-2} \ge 0 \implies g'(x) \ge g'_{min}(x)=g'(1)=0 \implies g(x)\ge g(1)=0 \implies f'(x)\ge 0 \implies f(x) \ge f(1)=0$
it remains 
$p\left(\dfrac{m-n}{2}\right)^2+mn \ge a^2+(p-1)b^2 \iff \dfrac{p(a^2+b^2)+(2-p)|a^2-b^2|}{2} \ge a^2+(p-1)b^2 \iff (a^2\ge b^2 \cap (a^2+(p-1)b^2 \ge a^2+(p-1)b^2)  , (a^2<b^2) \cap (b^2+(p-1)a^2 \ge a^2+(p-1)b^2 \iff (2-p)b^2 \ge (2-p)a^2)$ 
QED
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