TITLE: Show $(x+y)^a > x^a + y^a$ for $x,y>0$ and $a>1$
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: Note that the lhs and rhs are homogeneous hence we can make the assumption that $x + y = 1$ and then $0<x,y<1$ and so that $x^a<x$ and $y^a<y$ which leads to $$x^a + y^a < x + y = 1 = (x+y)^a$$