Almintas2l

2022-07-17

How to show $\left(a+b{\right)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in \left(0,1\right]$?
Does anyone happen to know a nice way to show that $\left(a+b{\right)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in \left(0,1\right]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)

We have
$1=\frac{a}{a+b}+\frac{b}{a+b}\le {\left(\frac{a}{a+b}\right)}^{n}+{\left(\frac{b}{a+b}\right)}^{n}$
for $0

Avery Stewart

The cases $a=0$ is trivial. For $a\ne 0$ let $m=1/n$ and $c={a}^{1/m}={a}^{n}$ and $d={b}^{1/m}={b}^{n}.$ Then
$\left(a+b{\right)}^{n}\le {a}^{n}+{b}^{n}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left({c}^{m}+{d}^{m}{\right)}^{1/m}\le c+d\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{c}^{m}+{d}^{m}\le \left(c+d{\right)}^{m}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}1+\left(d/c{\right)}^{m}\le \left(1+d/c{\right)}^{m}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(d/c{\right)}^{m}\le \left(1+d/c{\right)}^{m}-1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}m{\int }_{0}^{d/c}{x}^{m-1}dx\le m{\int }_{0}^{d/c}\left(1+x{\right)}^{m-1}dx$
which holds as $d/c\ge 0$ and $m-1\ge 0.$

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