Does this inequality hold true for all $\alpha\in\mathbb{R}$?

Question

Let $x$ and $y$ be two real and positive numbers. Let $\alpha\in\mathbb{R}$. I am trying to understand if the inequality $$ x^{\alpha} + y^{\alpha} \leq (x+y)^{\alpha}$$ holds true. By attemps, I found that it holds true only if $\alpha \geq1$. Could anyone please tell me if it is true?

Moreover, there is a simple way to justify it instead of proceeding by trial?

Thank you in advance!

Answer 1

If the inequality holds for all $x, y \geq 0$ the take $x=y=1$ to get $2 \leq 2^{\alpha}$ which implies $\alpha \geq 1$.

If $\alpha \geq 1$ consider the function $f(x)=(x+y)^{\alpha}-x^{\alpha}-y^{\alpha}$ for fixed $y$. Since $f'(x)=\alpha (x+y)^{\alpha -1} -\alpha x^{\alpha -1} \geq 0$ the function is increasing on $[0,\infty)$. Since $f(0)=0$ w get $f(x) \geq 0$ for all $x \geq 0$. This proves the inequality when $\alpha \geq 1$.

Answer 2

We have that

$$x^{\alpha} + y^{\alpha} \leq (x+y)^{\alpha} \iff \left(\frac x{x+y}\right)^\alpha+\left(\frac y{x+y}\right)^\alpha \le 1$$

and assuming wlog $x+y=1$

$$x^{\alpha} + y^{\alpha} \leq 1$$

which is true by convexity for $\alpha \ge 1$.

Answer 3

$f(x)=x^{\alpha}$ is a convex function for $\alpha\geq1$.

Let $x\geq y$.

Thus, $(x+y,0)\succ(x,y)$ and by Karamata we obtain: $$f(x+y)+f(0)\geq f(x)+f(y),$$ which gives $$(x+y)^{\alpha}\geq x^{\alpha}+y^{\alpha}.$$ For $0<\alpha<1$ the inequality is reversed.