Prove that $ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1$ holds

Question

Let $p_1, p_2>1$ and $a, b\in\mathbb{R}^+$ such that $a+b\geq2$. Could anyone please help me to prove that
$$ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1?$$
Here $p=\min(p_1, p_2)$. I am trying by using the inequality
$$(a+b)^p\leq 2^{p-1}(a^p + b^p),$$
but I am not able to obtain the desired result. Could anyone please help me or give me a hint?

Thank you in advance!

Answer 1

By assumption $a+b\ge2$. So at least one of these two numbers is larger than $1$. Assume $a\ge1$. Then $a^p\le a^{p_1}$. If $b\ge1$ then $b^p\le b^{p_2}$, if $b\in (0,1)$ then $b^p \le 1$. This proves $$ a^p + b^p \le a^{p_1} + b^{p_2} + 1. $$