# When does Equality for this Classical Inequality Hold?

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.

• Duplicate of your question: math.stackexchange.com/questions/143173/… Apr 28 at 3:29
• @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. Apr 28 at 3:46
• I understand - maybe you can use Newton's generalized binomial theorem (for real p) and some facts about binomial coefficients and determine when? Apr 28 at 4:53

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 $$a=b$$.
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 $$f=0.$$