When does Equality for this Classical Inequality Hold?

Question

When $a, b \in \mathbf{R}$ and $p \geq 1$, it is known that we have $$ |a + b|^p \leq 2^{p - 1}(|a|^p + |b|^p). $$ I am trying to see the sufficient and necessary condition of the equality of this inequality to hold.

My attempt is the following:

We wish to show that $$ |a + b|^p = 2^{p - 1}(|a|^p + |b|^p) $$ We start by noticing that this inequality turns into the triangle inequality when $p = 1$ and the equality for triangle inequality holds if and only if we have $a = cb$ for $c \in \mathbf{R}$. Now suppose this condition is true, we shall show if extra conditions are needed for general $p \geq 1$. Now with the condition $a = cb$ for $c \in \mathbf{R}$, we have $$ |a + b|^p = |cb + b|^p = |(c + 1)b|^p = |c + 1|^p |b|^p $$ On the other hand, we have $$ 2^{p - 1}(|a|^p + |b|^p) = 2^{p - 1}(|c|^p|b|^p + |b|^p) = 2^{p - 1}((|c|^p + 1)|b|^p) = 2^{p - 1}(|c|^p + 1)|b|^p. $$ That is, we need to have $$ 2^{p - 1}(|c|^p + 1) = |c+ 1|^p. $$ Therefore, we have if $a = cb$ for some $c \in \mathbf{R}$ such that $2^{p - 1}(|c|^p + 1) = (c + 1)^p$, then $|a + b|^p = 2^{p - 1}(|a|^p + |b|^p)$. However, I am not sure if this is a good enough condition to characterize the equality.

Answer 1

For $p>1$ the function $t \to t^{p}$ is striclty convex function on $[0,\infty)$. Hence, $|\frac {a+b} 2|^{p}\leq (\frac {|a|+|b|}2)^{p} <\frac {|a|^{p}+|b|^{p}}2$ (which is same as $ |a + b|^p < 2^{p - 1}(|a|^p + |b|^p). $) unless $a=b$. So equality holds only when $a=b$.

Answer 2

For $a\ge b\ge 0,$ with $a,b$ not both $0$: Let $a,b$ vary but with the restriction that $a+b=2m$ is constant. Let $f=2^{p-1}(a^p+b^p)-(2m)^p.$

Since $db/da=-1$ we have $df/da=p2^{p-1}(a^{p-1}-b^{p-1}),$ which is positive excspt when $a=b.$ Hence $f$ achieves a minimum only when $a=b=m,$ when $f=0.$