How to prove $x^m+y^m\leq c (x+y)^m$ with $x,y>0$ and $c>0$

Question

How to prove that for $m>1$ there exists a positive constant $c$ such that $x^m+y^m\leq c (x+y)^m$, where $x,y>0$?

Is it true for a "real" $m>1$ that

$$x\leq x+y \Rightarrow x^m \leq (x+y)^m \text{ and } y\leq x+y \Rightarrow y^m \leq (x+y)^m$$

then $x^m+y^m\leq 2(x+y)^m$.

Thank you

Answer 1

Actually $x^{m}+y^{m}\leq(x+y)^{m}$ for $x,y\geq 0$.

It suffices to prove for $1+x^{m}\leq(1+x)^{m}$.

Now let $\varphi(x)=(1+x)^{m}-(1+x^{m})$, obviously, $\varphi(0)=0$, and $\varphi'(x)=m(1+x)^{m-1}-mx^{m-1}=m((1+x)^{m-1}-x^{m-1})>0$, so $\varphi$ is increasing and hence $\varphi(x)\geq\varphi(0)=0$.

Answer 2

Let $x,y>0$, in particular $\frac yx>0$, and without loss of generality assume $x\geqslant y$. By the (generalized) Bernoulli Inequality we obtain for $m>1$ $$\left(1+\frac yx\right)^m\geqslant 1+m\frac yx\iff(x+y)^m\geqslant x^m+myx^{m-1}\geqslant x^m+my^m\geqslant x^m+y^m$$

$$\therefore~x^m+y^m\leqslant (x+y)^m$$

Therefore $c=1$, regardless of $x,y,m$. Anyway, as you were only asked to show that $x^m+y^m\leqslant c(x+y)^m$ holds for some positive constant $c$ your approach is perfectly fine aswell!