Is this inequality true? $ (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} $, for positive $x$ & $y$, and for $0 < \alpha < 1$ If $0 &lt; \alpha &lt; 1$, then 
  
  $$ (x + y)^{\alpha} &lt; x^{\alpha} + y^{\alpha} $$
  
  for $x$, $y$ positive.


Is this inequality true in general?

I tried using Young's Inequality: For $z,t &gt; 0$, and for $n$, $m$ such that $n+m=1$, then 

$$ z^n + t^m \leq nz + mt $$

So, using this we have 

$$ (x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)$$

which is not as tight as I want.
Hint:  try rewriting the inequality as
$$x^\alpha \left( 1 + t \right)^\alpha \le x^\alpha \left( 1 + t^\alpha \right)$$
where $t = y/x$.
Of course the case $\alpha=0$ or $\alpha=1$ is trivial, so suppose $0&lt;\alpha&lt;1$. Dividing both sides by $(x+y)^\alpha$ reduces the proof to the case $x+y=1$ (to see this, just do the substitution $x'=\frac{x}{x+y}, y'=\frac{y}{x+y}$). So suppose $x+y=1$. Then $0&lt;x&lt;1$, $0&lt;y&lt;1$, implying $x&lt;x^\alpha$ and $y&lt;y^\alpha$. Hence $$(x+y)^\alpha=1=x+y&lt;x^\alpha+y^\alpha.$$