Show $(x+y)^a > x^a + y^a$ for $x,y>0$ and $a>1$ This is a pretty straightforward question. I want to show $(x+y)^a > x^a + y^a$ for $x,y>0$ and $a>1$.
One way would be this. WLOG, suppose $x \leq y$. Then:
$(1+\frac{x}{y})^a >1+\frac{x}{y} \geq 1+(\frac{x}{y})^a$.
Noting that $\frac{1}{y^a}t_1>\frac{1}{y^a}t_2 \implies t_1 > t_2$, it follows that $(x+y)^a > x^a + y^a$.
But what other ways are there to show this inequality?

Write
$$ f(a) = \left( \frac{x}{x+y} \right)^a + \left( \frac{y}{x+y} \right)^a $$
It is clear that
$$ f(1) = \frac{x}{x+y} + \frac{y}{x+y} = 1 $$
It is also clear that
$$ f'(a) = \ln\left( \frac{x}{x+y} \right) \left( \frac{x}{x+y} \right)^{a-1} + \ln\left( \frac{y}{x+y} \right) \left( \frac{y}{x+y} \right)^{a-1} $$
whence
$$f'(a) < 0 $$
and therefore
$$ a > 1 \Rightarrow f(a) < 1$$
Thus
$$ \left( \frac{x}{x+y} \right)^a + \left( \frac{y}{x+y} \right)^a < 1 $$
or
$$\Big( x + y \Big)^a > x^a + y^a $$
