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I'm trying to prove the following inequality for $a$, $b$ $\in \mathbb{R}_{>0}$, $n$ $\in \mathbb{Z}_+$

$(a + b)^{\frac{1}{n}} \leq a^{\frac{1}{n}} + b^{\frac{1}{n}}$

At first glance I thought this would be provable by induction but had little luck so far. The next approach I've tried is to prove that the inverse holds instead as following

$(a + b)^{n} \geq a^{n} + b^{n}$

which is easily done using the the binomial theorem when $n$ is an integer but it feels like it needs to be done for all real $n$ for the argument to hold to prove the previous inequality.

ottizy
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  • See also https://math.stackexchange.com/q/1707969/9464 –  Nov 20 '17 at 02:05
  • $0 < (a+b)^{\frac 1n}$$ \le a^{\frac 1n} + b^{\frac 1n} $$\iff [(a+b)^{\frac 1n}]^n \le (a^{\frac 1n} + b^{\frac 1n})^n$$\iff a+b \le a+ b + \sum {n \choose i}a^{\frac in}b^{\frac {n-i}n} $$\iff \sum {n \choose i}a^{\frac in}$$b^{\frac {n-i}n} \ge 0$ – fleablood Nov 20 '17 at 04:04
  • @fleablood Really nice solution! Helped a alot. Thanks – ottizy Nov 20 '17 at 17:12

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