An triangle inequality $\sum_{i=1}^n b_i^\alpha \ge \sum_{i=1}^na_i^\alpha $ if $\alpha >1$

Question

Using my computer I discovered that:

if $a,b,c$ are sidelengths of a triangle, then

$(a+b-c)^\alpha+(b+c-a)^\alpha+(c+a-b)^\alpha \ge a^\alpha+b^\alpha+c^\alpha $ if $\alpha >1$

$(a+b-c)^\alpha+(b+c-a)^\alpha+(c+a-b)^\alpha \le a^\alpha+b^\alpha+c^\alpha $ if $0< \alpha <1$

My question: The generalization inequality as follow is corect?

Let and $\alpha>0$, $a_i>0$ for $i=\overline{1,n}$ and $S=a_1+a_2+....+a_n$. Let $b_i=S-(n-1)a_i \ge 0$ for $i=\overline{1,n}$, then

$\sum_{i=1}^n b_i^\alpha \ge \sum_{i=1}^na_i^\alpha $ if $\alpha >1$

$\sum_{i=1}^n b_i^\alpha \le \sum_{i=1}^na_i^\alpha $ if $0< \alpha <1$

Answer 1

This is immediate from Minkowski's inequality: for $p>1$, $$((x_1+y_1)^p+(x_2+y_2)^p+(x_3+y_3)^p)^{1/p}\leq (x_1^p+x_2^p+x_3^p)^{1/p}+(y_1^p+y_2^p+y_3^p)^{1/p}.$$ Let $x_1=y_2=a+b-c,x_2=y_3=b+c-a,x_3=y_1=c+a-b$ and $p=\alpha$. We then get $x_1+y_1=2a,x_2+y_2=2b,x_3+y_3=2c$ and hence $$2(a^\alpha+b^\alpha+c^\alpha)^{1/\alpha}\leq 2((a+b-c)^\alpha+(b+c-a)^\alpha+(c+a-b)^\alpha)^{1/\alpha},$$ $$a^\alpha+b^\alpha+c^\alpha\leq(a+b-c)^\alpha+(b+c-a)^\alpha+(c+a-b)^\alpha.$$ For $0<p<1$ the inequality in Minkowski reverses, so we get your second inequality.

For arbitrary $n$ we can repeat the reasoning by taking Minkowski for the sum of $n-1$ terms, which we take to be $b_2+\dots+b_n=na_1, b_1+b_3+\dots+b_n=na_2,\dots,b_1+\dots+b_{n-1}=na_n$. This gives $$n\left(\sum_{i=1}^n a_i^\alpha\right)^{1/\alpha}\leq n\left(\sum_{i=1}^n b_i^\alpha\right)^{1/\alpha}$$ for $\alpha>1$ and reverse inequality for $0<\alpha<1$.