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Does anyone happen to know a nice way to show that $(a+b)^n \le a^n+b^n$, where $a,b\geq 0$ 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 :)

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    $\begingroup$ Triangle Inequality. Look that up as a start (this is a generalization of the triangle inequality) $\endgroup$ Feb 18, 2017 at 18:21
  • $\begingroup$ If $a = 0$ or $b = 0$, equality is easy to see. If both are $> 0$, divide by $(a+b)^n$, then you want to show that $$\biggl(\frac{a}{a+b}\biggr)^n + \biggl(\frac{b}{a+b}\biggr)^n \geqslant 1.$$ Why is that the case for $0 < n \leqslant 1$? $\endgroup$ Feb 18, 2017 at 18:38

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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 $0<n\le 1.$

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  • $\begingroup$ Best one-liner this week. $\endgroup$ Aug 25, 2017 at 10:35
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Let $f(x)=x^n$, where $n\in(0,1]$ and $a\geq b$.

Hence, $f$ is a concave function and $(a+b,0)\succ(a,b)$.

Thus, by Karamata $$(a+b)^n+0^n\leq a^n+b^n$$ and we are done!

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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 $$(a+b)^n\leq a^n+b^n\iff (c^m+d^m)^{1/m}\leq c+d\iff$$ $$\iff c^m+d^m\leq (c+d)^m\iff 1+(d/c)^m\leq (1+d/c)^m\iff$$ $$\iff (d/c)^m\leq (1+d/c)^m-1\iff$$ $$\iff m\int_0^{d/c}x^{m-1}dx\leq m\int_0^{d/c}(1+x)^{m-1}dx$$ which holds as $d/c\geq 0$ and $m-1\geq 0.$

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  • $\begingroup$ I always do it the hard way. $\endgroup$ Aug 25, 2017 at 10:36
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Assume that $a \ge b$. If $a = b = 0$, the result is immediate.

If $a > 0$, divide $(a+b)^n \le a^n+b^n $ by $a^n$ to get $(1+b/a)^n \le 1+(b/a)^n $.

Since $b \le a$, $0 \le b/a \le 1$, so this becomes $(1+x)^n \le 1+x^n $ where $x = b/a$.

Let $f(x) =1+x^n-(1+x)^n $.

$f(0) = 0$ and $f(1) =2-2^n \ge 0 $ since $0 < n \le 1$.

$f'(x) =nx^{n-1}-n(1+x)^{n-1} =n(x^{n-1}-(1+x)^{n-1}) =n(\frac1{x^{1-n}}-\frac1{(1+x)^{1-n}}) $.

Since $0 < n \le 1$, $1-n \ge 0$ so that $x^{1-n} \le (1+x)^{1-n} $ so that $\frac1{x^{1-n}}\ge\frac1{(1+x)^{1-n}} $ so that $f'(x) \ge 0$.

Since $f(0) = 0$ and $f'(x) \ge 0$ for $0 < x\le 1$, $f(x) \ge 0 $ for $0 \le x \le 1$.

Note that the inequality goes the other way if $n > 1$.

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