Triangle inequality raised to fractional powers

Question

Does the inequality $|x + y|^\alpha \leq |x|^\alpha + |y|^\alpha$ where $\alpha \in [0, 1)$, hold for all $x,y$ ??

Answer 1

Yes. First, notice that it suffices to prove it for $x,y \geq 0$. Indeed, if we know $(|x|+|y|)^\alpha \leq |x|^\alpha + |y|^\alpha$, but $|x+y|^\alpha \leq (|x|+|y|)^\alpha$ since $\alpha > 0$, giving the required general result.

Next, notice that it is trivially true for $x = 0$, so we only need to prove it for $x>0$, $y \geq 0$.

But now we must show $(x+y)^\alpha \leq x^\alpha + y^\alpha$, so divide through by the (non-zero) $x$, giving the equivalent statement that $(1 + y/x)^\alpha \leq 1 + (y/x)^\alpha$. So, we only need to show $(1+t)^\alpha \leq 1 + t^\alpha$ for $t \geq 0$.

For this, we can consider the function $f(t) = (1+t)^\alpha - t^\alpha$, and our goal is to say that $f(t) \leq 1$ for $t \geq 0$. To this end, observe $f(0) = 1$, and $f'(t) = \alpha [(1+t)^{\alpha-1} - t^{\alpha - 1}]$, and if we can show $f'(t) \leq 0$ for $t \geq 0$, this would imply $f$ is decreasing, so $f(t) \leq 1$ as required.

But $t \mapsto t^{\alpha - 1}$ is a decreasing function since $\alpha < 1$, so $(1+t)^{\alpha-1} - t^{\alpha-1} < 0$, giving the result.

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This is ultimately very similar to Youem's proof, but the above is my exact thought process for proving this, and may help intuition. For homogeneous inequalities like these, dividing through to give a function we can analyse using simple calculus is often a useful trick.

Answer 2

For the $x\ge 0,y\ge 0$. $f : t \mapsto (t + x)^{\alpha} - t^{\alpha}$ is decreasing function on $[0,+\infty)$ (take the derivative $f'(t) = \alpha ((t+x)^{\alpha -1} - t^{\alpha - 1}) < 0$ since $\alpha < 1$) so $$|x|^{\alpha} = f(0) \ge f(y) = |x+y|^\alpha - y^\alpha$$