The following formula is given:
$(a+b)^\alpha \leq a^\alpha+b^\alpha$
What is the minimum $0<\alpha<1$ somehow the above inequality is being satisfied for all $a,b > 0$?
Intuitively as if $\alpha \rightarrow 0$, the above inequality is satisfied, for the set of solutions of the question, there is no any infimum. Is it correct?
Also, I've heard that the above inequality is true based on the above constraints, but I am seeking for a formal proof.
Define $f(a) = a ^ \alpha + b ^ \alpha - (a+b)^\alpha$. So,
$$ f'(a) = \alpha a ^{\alpha - 1} - \alpha (a+b)^{\alpha-1} = \alpha (a ^{\alpha - 1} - (a+b)^{\alpha-1})= \alpha (\frac{1}{a ^{1-\alpha}} - \frac{1}{(a+b)^{1-\alpha}}) > 0$$
so, $f$ is increasing for $a$ and $b$. As $a,b > 0$ and for $a = 0$ and $b = 0$, we have $a ^ \alpha + b ^ \alpha - (a+b)^\alpha = 0$, so the phrase is a non-negative function and the proof is completed.