Almintas2l

2022-07-17

How to show $(a+b{)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in (0,1]$?

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

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

slapadabassyc

Beginner2022-07-18Added 21 answers

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 $0<n\le 1.$

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

for $0<n\le 1.$

Avery Stewart

Beginner2022-07-19Added 2 answers

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

$(a+b{)}^{n}\le {a}^{n}+{b}^{n}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({c}^{m}+{d}^{m}{)}^{1/m}\le c+d\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{c}^{m}+{d}^{m}\le (c+d{)}^{m}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}1+(d/c{)}^{m}\le (1+d/c{)}^{m}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(d/c{)}^{m}\le (1+d/c{)}^{m}-1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}m{\int}_{0}^{d/c}{x}^{m-1}dx\le m{\int}_{0}^{d/c}(1+x{)}^{m-1}dx$

which holds as $d/c\ge 0$ and $m-1\ge 0.$

$(a+b{)}^{n}\le {a}^{n}+{b}^{n}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({c}^{m}+{d}^{m}{)}^{1/m}\le c+d\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{c}^{m}+{d}^{m}\le (c+d{)}^{m}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}1+(d/c{)}^{m}\le (1+d/c{)}^{m}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(d/c{)}^{m}\le (1+d/c{)}^{m}-1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}m{\int}_{0}^{d/c}{x}^{m-1}dx\le m{\int}_{0}^{d/c}(1+x{)}^{m-1}dx$

which holds as $d/c\ge 0$ and $m-1\ge 0.$

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