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?
 A: You can use elementary calculus to prove this: Let $x_0>0$ be fixed, and consider the function $f:[0,\infty)\to\mathbb{R}$, $f(y)=(x_0+y)^a-y^a$. The derivative of $f$ is $f'(y)=a\left((x_0+y)^{a-1}-y^{a-1}\right)$. Since $a>1$ and $x_0,y\geq 0$, then $f'>0$, so $f$ is strictly increasing. Then, for every $y\in\mathbb{R}$, $f(y)>f(0)$, and this is equivalent to $(x_0+y)^a>x_0^a+y^a$.
A: 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
$$\Big( x + y \Big)^a > x^a + y^a $$
A: 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 $$x^a + y^a < x + y = 1 = (x+y)^a$$
A: Let $x\geq y$  and  $t=\frac{x}{y}\ \ \text{and} f(t)=(1+t)^a-(1+t^a)$ since $f'(t)=a(1+t)^{a-1}-at^{a-1}>0$ therefore$f(t)>f(0) $ and so we have $f(t)=(1+t)^a-(1+t^a)> 0$ thus $(1+t)^a>(1+t^a) $ this show that $(1+\frac{x}{y})^a>1+(\frac{x}{y})^a$ all together we see that $(x+y)^a > (x)^a+(y)^a$
A: Please tell me what's wrong with this:
On the one hand, we have $(x+y)^a$ equaling the sum of several positive numbers. I am referring to the binomial theorem. On the other hand, $x^a$ and $y^a$ are just two of those several positive numbers. So, $(x+y)^a > x^a + y^a$.
Again, I am just using the binomial theorem to show that $(x+y)^a$ is equal to an expression which contains $x^a, y^a$ and at least 1 other positive, nonzero term. 
(Yes, this is poorly explained, but I only have a few minutes to type this up. I've never posted before and I don't know how to make the pretty formatting)
If someone knows what I'm trying to say, and sees whats wrong with my thinking, let me know.
A: Hint: Using the Pockhammer symbol in the binomial expansion,  $$\begin{align}(x+y)^a&=\sum_{n=0}^{\infty}\frac{(a)_n}{n!}x^{a-n}y^n \\
&>\frac{(a)_0}{0!}x^a+\frac{(a)_{\lfloor a\rfloor}}{(\lfloor a\rfloor)!}x^{a-\lfloor a\rfloor}y^{\lfloor a\rfloor} \\
&\stackrel{?}{>}x^a+y^a,\end{align}$$ where $\lfloor a\rfloor$ is the floor of $a$.
