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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.

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  • $\begingroup$ Duplicate of your question: math.stackexchange.com/questions/143173/… $\endgroup$
    – Kyler S
    Apr 28 at 3:29
  • $\begingroup$ @KylerS Thank you for your comment. However, if I read it correctly, I do not think the post answers my questions as to the conditions when the equality of this inequality holds. $\endgroup$ Apr 28 at 3:46
  • $\begingroup$ I understand - maybe you can use Newton's generalized binomial theorem (for real p) and some facts about binomial coefficients and determine when? $\endgroup$
    – Kyler S
    Apr 28 at 4:53

2 Answers 2

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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 $a=b$.

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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 $f=0.$

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