how to prove $a+b< (a^{\alpha}+b^{\alpha})^{\frac{1}{\alpha}}$

Question

how to prove $a+b< (a^{\alpha}+b^{\alpha})^{\frac{1}{\alpha}}$ when $0<\alpha<1$.(Im not suure if is for every a,b) I am really dont try anything, but i think that maybe is using integration in some region, i will appreciate if you give me one hint, thank you so much.

why somebody put -1, is really easy? i wrote i dont have idea how to star — weymar andres, Dec 29 '20 at 13:49
Look up subadditive functions, and why concave functions are subadditive. The function $x \to x^{\alpha}$ is concave for $0<\alpha<1$. — Sarvesh Ravichandran Iyer, Dec 29 '20 at 13:49
maybe i understand, is becasuse $f(x)=x^{\alpha}$ imply $f''(x)=\alpha (\alpha -1)x^{\alpha-2}<0$ — weymar andres, Dec 29 '20 at 13:55
That proves it is concave. From concavity to subadditivity is a small step, and subadditivity , upon rearrangement, gives you that statement there. — Sarvesh Ravichandran Iyer, Dec 29 '20 at 13:56
im confuse because, $f$ is concave then $f(tx+(1-t)y)>tf(x)+(1-t)f(y)$ when i choose $t=\frac{1}{2}$ we have $ (\frac{x+y}{2})^{\alpha}>\frac{1}{2}(x^{\alpha}+y^{\alpha})$ — weymar andres, Dec 29 '20 at 14:09
That is not a contradiction, right? You have $2^{\alpha}$ on one side and $2$ on the other side, so this inequality is true, and subadditivity is true. — Sarvesh Ravichandran Iyer, Dec 29 '20 at 15:04

score 1 · Accepted Answer · answered Dec 29 '20 at 15:58

You can prove it directly just using strict monotonicity of an appropriate function.

First note that you are dealing with an inequality among real numbers and there expressions like $a^{\alpha}, b^{\alpha}$ with real $0<\alpha<1$ are defined for $a,b\geq 0$.

Your inequality is wrong if $a=0$ or $b=0$. So, you may want to prove it for $a,b>0$.

Without loss of generality we can assume that $0<b\leq a$ and set $x=\frac ba$: $$a+b < \left(a^{\alpha}+ b^{\alpha}\right)^{\frac 1{\alpha}}\Leftrightarrow (a+b)^{\alpha} < a^{\alpha}+ b^{\alpha}$$ $$\stackrel{x=\frac ba}{\Leftrightarrow}(1+x)^{\alpha} < 1+x^{\alpha} \text{ for } 0<x\leq 1$$

The function $f(x) = 1+x^{\alpha} - (1+x)^{\alpha}$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Furthermore you have $f(0) = 0$. So, it is enough to show that $f'(x) >0$ on $(0,1)$:

$$f'(x) = \alpha \left(\underbrace{\frac 1{x^{1-\alpha}}}_{>1 \text{ on } (0,1)} - \underbrace{\frac 1{(1+x)^{1-\alpha}}}_{<1 \text{ on } (0,1)}\right) > 0$$

how to prove $a+b< (a^{\alpha}+b^{\alpha})^{\frac{1}{\alpha}}$

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