A generalization of this type of mean to power > 2

Question

Let $a,b \in (0, \infty)$ and $0 < \theta < 1$. Then $$ (a + b)^2 \leq \frac{1}{\theta} a^2 + \frac{1}{1 - \theta} b^2. $$

Can you generalize to a power $m \geq 2$?

(The question is not whether the inequality holds with $2$ replaced by $m$.)

Answer 1

You can generalise using Holder's inequality that: $$\left(\frac1\theta a^m + \frac1{1-\theta}b^m\right)(\theta+\overline{1-\theta})(1+1)^{m-2} \ge (a+b)^m \implies \frac1\theta a^m + \frac1{1-\theta}b^m \ge \frac{(a+b)^m}{2^{m-2}}$$

Answer 2

This is a different generalization: $$ (a+b)^m\le\frac{1}{\theta^{m-1}}\,a^m+\frac{1}{(1-\theta)^{m-1}}\,b^m. $$ It follows from the inequality $$ (1+x)^m\le\frac{1}{\theta^{m-1}}+\frac{1}{(1-\theta)^{m-1}}\,x^m,\quad0<x<1. $$ To prove it, minimize for $x$ fixed as a function of $\theta$ the right hand side. The minimum is attained at $\theta=1/(1+x)$ and its value is precisely $(1+x)^m$.

Answer 3

Just for completeness, combining the two generalisations we have, for $2 \le n \le m$ we can show by Holder's inequality: $$\left( \frac{a^m}{\theta^n} + \frac{b^m}{(1-\theta)^n} \right)(\theta +\overline{1-\theta})^n(1+1)^{m-n-1} \ge (a+b)^m$$

$$\implies \frac{a^m}{\theta^n} + \frac{b^m}{(1-\theta)^n} \ge \frac{(a+b)^m}{2^{m-n-1}}$$ Equality is when $a=b=1, \; \theta=\frac12$.