A simple inequality: $(x+y)^p>x^p+y^p?$ Let $x,y&gt;0$ and $p&gt;1$. Is it always true that $$(x+y)^p&gt;x^p+y^p?$$

I think it is true. But can not prove it.
Since the inequality is homogeneous, we may assume without loss of generality that $y=1$ and $x\in(0,1]$, then prove:
$$ \forall x\in(0,1],\qquad (1+x)^p &gt; 1+x^p.\tag{1}$$
On the other hand, if we set $f(x)=(1+x)^p-x^p$, it is straightforward to check that $f'(x) = (p-1)\cdot\left((1+x)^{p-1}-x^{p-1}\right)&gt;0$, hence $f$ is increasing on $(0,1]$. Since $f(0)=1$, $(1)$ is proved.
Let $x,y&gt;0$ and $p&gt;1$ so we must prove $(x+y)^p &gt; x^p + y^p$. We can prove this by saying that $(x+y)^p = x^p + y^p + c(x,y)$ where $c(x,y)$ are terms of $x,y$ that are combined. If $x,y&gt;0$ then so must $c(x,y)$ because it will be the sum of products of positive numbers. Example for $c(x,y)$ $(x+y)^2 = x^2 + y^2 + 2xy$. Thus as we know $c(x,y)&gt;0$ we can say $(x+y)^p = x^p + y^p + c(x,y) &gt; x^p + y^p$ and if we subtract $x^p + y^p$ from both sides we get $c(x,y)&gt;0$ which is true.