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If $0 < \alpha < 1$, then

$$ (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} $$

for $x$, $y$ positive.

Is this inequality true in general?

I tried using Young’s Inequality: For $z,t > 0$, 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.

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Hint: try rewriting the inequality as

$$x^\alpha \left( 1 + t \right)^\alpha \le x^\alpha \left( 1 + t^\alpha \right)$$

where $t = y/x$.

Let $x\geq y$.

Hence, $(x+y,0)\succ(x,y)$ and since $f(x)=x^{\alpha}$ is a concave function, by Karamata we obtain:

$$(x+y)^{\alpha}=(x+y)^{\alpha}+0^{\alpha}\leq x^{\alpha}+y^{\alpha}$$

and we are done!

LEt $f(t) = 1 + t^{\alpha} – (1+t)^{\alpha}$, $t > 0$ and $0 < \alpha < 1 $. We have

$$ f'(t) = \alpha t^{\alpha – 1} – \alpha(1+t)^{\alpha-1} = \alpha t^{\alpha-1} \left( 1 – \left( \frac{ 1 + t }{t} \right)^{\alpha-1} \right)$$

Since $\frac{1+t}{t} > 1$ and $\alpha – 1 <0$, then $1 – \left( \frac{ 1 + t}{t} \right)^{\alpha – 1} > 0$. Therefore,

$$ f'(t) > 0 \; \text{all} \; t>0 \implies f(t) > f(0) \implies 1 + t^{\alpha} > (1+t)^{\alpha}$$

Assume $x,y > 0$, then $t = \frac{y}{x} > 0$ and thus

$$ 1 + (y/x)^{\alpha} > ( 1 + y/x)^{\alpha} \implies x^{\alpha} + y^{\alpha} > (x+y)^{\alpha}$$

$$(x+y)^a\lt x^a+y^a\iff x^a\left(1+\dfrac yx\right)^a=x^a\left(1+(\dfrac yx)^a\right)$$ Hence it is enough to prove

$$(1+x)^a\lt 1+x^a$$

The function $$f(x)=1+x^a-(1+x)^a$$ has the derivative $$f'(x)=a(x^{a-1}-(1+x)^{a-1})$$ which is positive for $0\lt a\lt 1$ and $x$ positive because

$$a(x^{a-1}-(1+x)^{a-1})\gt 0\iff \frac{1}{x^{1-a}}-\frac{1}{(1+x)^{1-a}}\gt 0$$ Consequently $f(x)$ is increasing for $x\gt 0$ so because $f(0)=0$ we have always

$$f(x)\gt 0\iff(1+x)^a\lt 1+x^a$$

A hint that I like for problem solving or maybe a different way to conceptualize these problems. Consider these two expressions as functions of 3 variables, $x,y,\alpha$. In one variable calculus you can use derivative to find maximum points or minimums. You can use generalized methods here. Consider term one subtracted from term two. If it has a global minimum that says something related to this inequality.

Of course the case $\alpha=0$ or $\alpha=1$ is trivial, so suppose $0<\alpha<1$. 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 $0<x<1$, $0<y<1$, implying $x<x^\alpha$ and $y<y^\alpha$. Hence $$(x+y)^\alpha=1=x+y<x^\alpha+y^\alpha.$$

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