A binomial expansion inequality.

Question

Let us consider $x, y > 0$ then given $n \in \mathbb{N}$ the following inequality holds true $$ (x+y)^\frac{1}{n} \leq x^\frac{1}{n} + y^\frac{1}{n} $$ since $$ ((x+y)^\frac{1}{n})^n = x + y \leq (x^\frac{1}{n} + y^\frac{1}{n})^{n} = x+ y +\text{some positive terms}. $$ Now let us consider $x, y$ as above and let $\alpha \in (0,1)$ then does the following inequality holds true, and any suggestion to prove it? $$ (x+ y)^\alpha \leq x^\alpha + y ^\alpha. $$

Answer 1

$$ (x+y)^\alpha-x^\alpha=\int_x^{x+y}\alpha t^{\alpha-1}\,dt\le \int_{x}^{x+y}\alpha (t-x)^{\alpha-1}\,dt=y^\alpha-\require{cancel}\cancel{0^\alpha}. $$

Answer 2

Also, you can use Karamata here.

Indeed, let $f(x)=x^{\alpha}$ and $x\geq y$.

Thus, $f$ is a concave function and $(x+y,0)\succ(x,y).$

Id est, by Kartamata $$f(x+y)+f(0)\leq f(x)+f(y),$$ which is your inequality.