Prove $a^{1/k}+b^{1/k}\gt (a+b)^{1/k}$

Question

Prove the statement(#):

(#) Suppose $k$ is an integer greater than $1$. Suppose $a,b$ are positive real numbers. Then $a^{1/k}+b^{1/k}\gt (a+b)^{1/k}$.

How should I start the proof? Any inequalities I can apply? Which topic is related to this proof? Plz give me some hints:( Thanks.

What happens when you expand $(LHS)^{k}$ using Binomial Theorem? — Kavi Rama Murthy, May 05 '22 at 09:56
Let divide both parts on $a^{1/k}$, then we need only prove $1+x^{1/k}>(1+x)^{1/k}$ — Ivan Kaznacheyeu, May 05 '22 at 10:46

score 2 · Accepted Answer · answered May 05 '22 at 10:26

Firstly whenever proving any sort of algebraic (in)equality I would try to isolate a useful property to simplify the equation. In this case,

$$ (a^{\frac{1}{k}}+b^{\frac{1}{k}})^{k} > [(a+b)^{\frac{1}{k}}]^{k} \\ (a^{\frac{1}{k}}+b^{\frac{1}{k}})^{k} > a+b $$

as both sides have an exponent $1/k$.

Secondly, I would try to identify any useful properties that I know something about. So in this case, $(p+q)^{k}$ is a very common binomial form.

Then from this point onwards the proof is pretty straightforward. I would recommend always starting with this sort of approach. Sometimes you'll get stuck because novel ideas are required but it'll at least help you get started with most proofs of this form.

Prove $a^{1/k}+b^{1/k}\gt (a+b)^{1/k}$

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