Advanced triangle inequality

Question

This problem came up in a series of Taylor series exercises and I have no idea where the correlation could be. Let $a \in (0,1)$ and $u,v \in \mathbb{R}_{+}$, prove

$$ (u+v)^{a} \leq u^{a}+v^{a} $$

I have tried several things. The Binomial theorem is not applicable since the exponent is rational. I tried rewriting $a=1/x$ where $x \in {x \in \mathbb{N}\ \{0,1\}}$ so we get:

$$ (u+v)^{1/x} \leq u^{1/x}+v^{1/x} $$ which just makes it clearer that it is about roots but doesn't do much to solve it. Tricks like using the third binomial theorem are also not applicable as well, as they only work for square roots. I tried shifting around the terms a little bit, but it doesn't seem to do much.

I'm thinking, I'm supposed do something more "profound" than an elementary proof, but I have nowhere to start. I'd appreciate a hint on where to go from here

What I did in the exercise before( maybe it's relevant) : I showed for example, that $f_{a}:(-1,1)\rightarrow \mathbb{R}, x \mapsto (1+x)^{a}$ can be approximated at $x_0=0$ with the taylor series and that they are equal at that point.

Answer 1

The fact follows from basic properties of exponentiation. Recall that if $0

With this in mind, $$1=\frac{u}{u+v}+\frac{v}{u+v}\le \left (\frac{u}{u+v}\right )^a+\left (\frac{v}{u+v}\right )^a=\frac{u^a+v^a}{(u+v)^a}$$ which is equivalent to our claim.

Answer 2

For $0<\alpha<1$ and $\dfrac120$ that shows $f(x)0$ and $q>0$ with $p+q=1$ then $$p^\alpha+q^\alpha\geq1$$ let $p=\dfrac{u}{u+v}$ and $q=\dfrac{v}{u+v}$ so $p+q=1$ and $$(\dfrac{u}{u+v})^\alpha+(\dfrac{u}{u+v})^\alpha\geq1$$ then $$(u+v)^\alpha\leq u^\alpha+v^\alpha$$