Prove that $ba^k + ab^k \leq a^{k+1} + b^{k + 1}$ for $a, b \geq 0$, $k \in \mathbb N$

Question

I'm in the middle of a proof by induction that for all $n \in \mathbb N$, $\left(\frac{a + b}{2}\right)^n \leq \frac{a^n + b^n}{2}$ for all nonnegative real numbers $a, b$. I've reached a point in my proof where I want to show that $ba^k + ab^k \leq a^{k + 1} + b^{k + 1}$, and I'm stuck here.

I attempted to prove this by showing a more general statement that $c + d \leq \frac{a}{b}c + \frac{b}{a}d$ for all $a, b > 0$ and $c, d \geq 0$, but I quickly realized that I can't prove it if it isn't true (whoops). That was my only idea; now I'm stuck. Well, not really -- my only other idea was to prove this statement itself by induction, but I'm convinced there's another way to show it without having a nested proof by induction. What must I consider to discover this route?

Thank you very much in advance.

However, if $a\geq b \geq 0$ and $c \geq d \geq 0$, then $(a-b)(ca-db) \geq 0$. — Catalin Zara, Jul 27 '20 at 00:25
Does this answer your question? Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$ — Martin R, Jul 27 '20 at 06:51

score 3 · Accepted Answer · answered Jul 27 '20 at 00:48

3

Hint: For $a,b\geq 0,$ and $k\in{\mathbb N},$ $$(a-b)(a^k-b^k)\geq 0.$$

answered Jul 27 '20 at 00:48

Pythagoras

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Prove that $ba^k + ab^k \leq a^{k+1} + b^{k + 1}$ for $a, b \geq 0$, $k \in \mathbb N$

1 Answers1