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I don't really know how to start here.

I tried to proof it by induction but don't know how to continue the inductive step. I got $ (a+b)^{1/(p+1)} \leq a^{1/(p+1)} + b^{1/(p+1)} $

I also tried to figure out a way by ignoring the $^{1/n}$ and to write $(a+b)^c \leq a^c + b^c$ instead, but couldn't figure out how to continue without using proof by induction because $c$ would be a rational number.

So does anyone have an idea on how to solve this? Which method to use?

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  • $\begingroup$ Binomial theorem? $\endgroup$ – ÍgjøgnumMeg Nov 21 '16 at 0:19
  • $\begingroup$ should I still use proof by induction? $\endgroup$ – Lukas Nov 21 '16 at 0:34
  • $\begingroup$ Use Jensen's inequality $\endgroup$ – Anon Nov 21 '16 at 1:07
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I assume $a,b\geq 0$. The binomial formula implies: $(a^{1/n}+b^{1/n})^n=(a^{1/n})^n+(b^{1/n})^n+..=a+b+...\geq a+b=((a+b)^{1/n})^n$

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As has been mentioned before, the Binomial Theorem works well here. The binomial theorem says that you can expand $(a+b)^n$ in the following way: $$(a+b)^n = \sum_{i = 0}^n\binom{n}{i}a^ib^{n-i}$$ The easiest way to apply this is to prove the inequality for $a+b\leq(a^{1/n}+b^{1/n})^n$. To do this, we apply Binomial Theorem to the right hand side: $$(a^{1/n}+b^{1/n})^n = \sum_{i = 0}^n \binom{n}{i}(a^{1/n})^i{(b^{1/n})}^{n-i}$$ Importantly, this sum on the right hand side is: $$\sum_{i = 0}^n \binom{n}{i}(a^{1/n})^i{(b^{1/n})}^{n-i} = b+\sum_{i = 1}^{n-1} \binom{n}{i}(a^{1/n})^i{(b^{1/n})}^{n-i}+a = a+b+c$$ Where $c$ is the rest of the sum. Assuming that $a,b\geq 0$, we'll get that $c\geq 0$ too, so $a+b+c\geq a+b$. But we have from before that $a+b+c = (a^{1/n}+b^{1/n})^n$, so this is: $$(a^{1/n}+b^{1/n})^n\geq a+b\implies a^{1/n}+b^{1/n}\geq (a+b)^{1/n}$$

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  • $\begingroup$ Thank you for the answer, I understand it but got one question. What happens if n = {1,2,3,4,...}? $\endgroup$ – Lukas Nov 23 '16 at 23:39
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Let $0<c<1$. Define $$ f(x)=(a+x)^{c}-a^{c}-x^{c}, x>0. $$ Then $$ f'(x)=c(\frac{1}{(a+x)^{1-c}}-\frac{1}{x^{1-c}})<0, x>0$$ and hence $f(x)$ is decreasing for $x\ge0$. Thus $f(x)\le f(0)=0$ for any $x\ge0$ and hence for $b\ge0$, $f(b)\le 0$, or $$ (a+b)^{c}\le a^{c}+b^{c}. $$

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Here we prove a more general result.

For any $0<p<1$ and $a\geqslant 0,b\geqslant 0$, there is $$ (a+b)^p\leqslant a^p+b^p\tag1 $$ If $a=b=0$, $(1)$ clearly holds. Assume $a\neq 0$ and let $x=\frac{a}{a+b}$. Then $(1)$ becomes $$ 1\leqslant x^p+(1-x)^p\tag2 $$ Clearly $0\leqslant x\leqslant 1$. Let $f(x)=x^p+(1-x)^p$. Then $$ f'(x)=px^{p-1}-p(1-x)^{p-1} $$ Since $f'(1/2)=0$ and \begin{cases} f'(x)>0, \quad 0<x<1/2\\ f'(x)<0, \quad 1/2<x<1 \end{cases} $f(1/2)$ is the only maximum point in $[0,1]$. Since $f(0)=f(1)=1$, $f(x)\geqslant 1$ on $[0,1]$. And so $(1)$ is true.

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