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I'm trying to show the following:

If $x,y \geq 0$, then $(x+y)^k \geq x^k + y^k$ for all $k \in \mathbb{R_{\geq 1}}$

So, far I've tried a few things, but nothing seems to stick. Clearly $x \leq x+ y$ and since both sides of the inequality are positive, the inequality $x^k \leq (x + y)^k$ will hold for $k \geq 1$. Similarly, $y^k \leq (x+y)^k$. Adding both inequalities together, we obtain: $x^k + y^k \leq 2(x+y)^k$. While this is close, of course, it is not what we want to show.

Alternatively, I was thinking that if we fix $x,y \geq 0$, let $f(k) = (x+y)^k$, and let $g(k) = x^k + y^k$, we can show using the binomial theorem that $(x+y)^r \geq x^r + y^r$ for all $r \in \mathbb{Z^+}$. Then, if we can show both $f$ and $g$ intersect only when $k =1 $, we might have better luck proving the statement since then we would have $(x+y)^r > x^r + y^r$ for all positive integers $r \geq 2$. A real number $m \notin \mathbb{Z}$ with the property that $f(m) = g(m)$ could not then exist since then that would violate the fact that $k=1$ is the only intersection. We could then invoke the continuity of $f$ and $g$ together with the fact that $(x+y)^r > x^r + y^r$ for all positive integers $r \geq 2$ to obtain our result. Of course, all of this is dependent on rigorously showing that $f$ and $g$ intersect only when $k=1$.

Otherwise, I'm running low on ideas. Any hints would be greatly appreciated.

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$(x+y)^{k}-x^{k}$ is an increasing function of $x$ because its derivative $k(x+y)^{k-1}-kx^{k-1} \geq 0$. At $x=0$ the value is $y^{k}$. Hence $(x+y)^{k}-x^{k} \geq y^{k}$.