Prove that $ a^{k} + b^{k} + c^{k} \ge a^{k-1}b + b^{k-1}c + c^{k-1}a$

Question

Prove that $ a^{k} + b^{k} + c^{k} \geq a^{k-1}b + b^{k-1}c + c^{k-1}a$ where $a, b, c$ are positive real numbers and $k$ is a positive integer.

I don't know how to start.

Answer 1

Hint:

Use Cauchy-Schwarz inequality with the vectors $(a^{k-1}, b^{k-1}, c^{k-1})$ and $(u,v,w)$. When do we have equality?

B.t.w., the strict inequality is false if $a=b=c$.

Answer 2

Since $\left(a^{k-1},b^{k-1},c^{k-1}\right)$ and $(a,b,c)$ are the same ordered,

your inequality follows from Rearrangement: $$a^k+b^k+c^k=a^{k-1}\cdot{a}+b^{k-1}\cdot{b}+c^{k-1}\cdot{c}\geq a^{k-1}b+b^{k-1}c+c^{k-1}a$$ Done!

Answer 3

A natural approach is the following.

Rewrite the inequality into $$a^{k-1}(a-b) + b^{k-1}(b-c) + c^{k-1}(c-a) \ge 0.$$ Notice that $c-a = -(a-b) - (b-c)$, the last inequality can be rewritten as $$a^{k-1}(a-b) + b^{k-1}(b-c) - c^{k-1}(a-b) - c^{k-1}(b-c) \ge 0.$$ Regrouping: $$(a^{k-1} - c^{k-1})(a-b) + (b^{k-1}- c^{k-1})(b-c) \ge 0.$$ The first term is non-negative if we assume (WLOG) $a=\max(a,b,c)$, the second term is always non-negative, thus the inequality is clearly true.