Is it possible to show that $$x+y\geq x^\alpha y^{1-\alpha}$$ for $\alpha \in(0,1]$ and $x,y\in[0,\infty)$?
I tried to manipulate it algebraically, but it does not give me any anything. Equivalently, we need to show that $$x^\alpha - y^\alpha \leq xy^{\alpha - 1}$$
$$x+y\geqslant\max\{x,y\}=\max\{x,y\}^\alpha\max\{x,y\}^{1-\alpha}\geqslant x^\alpha y^{1-\alpha}$$
If $y = 0$ this is obvious for any $x \geq 0$.
If $y \neq 0$ this is equivalent to:
$$\frac{x}{y} + 1 \geq \frac{x^\alpha}{y^\alpha}$$
Taking $z = \frac{x}{y}$ we need to prove that
$$z+1\geq z^\alpha$$
For $\alpha\in (0,1]$ and $z\geq 0$.
Now we have two cases: $z \geq 1$ and $0\leq z < 1$.
In the first case, the function $z^t$ is increasing so $z^\alpha \leq z^1$ since $\alpha \leq 1$ so $z^\alpha \leq 1 + z$.
In the second case, the function $z^t$ is decreasing so $z^\alpha < z^0 = 1$ since $\alpha > 0$ so $z^\alpha < 1 + z$
