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$ n \in \mathbb{N},\ n\ge 2$
$ c_1, c_2, a, b \in \mathbb{R} $

For $ \forall a,b $ determine the maximum value of $ c_1 $ and the minimum value of $ c_2 $ that makes true the following inequation:

$ c_1(a+b)\le(\sqrt[\leftroot{-1}\uproot{3}n]{a} + \sqrt[\leftroot{-1}\uproot{3}n]{b})^n \le c_2(a+b) $

My doubts are:
1° there are two variables, how can I maximize $ c_1 $ if depends from $ a, b $ ?
A my idea was to set a function $ c_1 = f(a,b) = \frac{(\sqrt[\leftroot{-1}\uproot{3}n]{a} + \sqrt[\leftroot{-1}\uproot{3}n]{b})^n}{a+b} $, and maximize that function, but I don't think it's the right way...
2° Since $ n $ is any number, I can't expand the binomial and simplify the expression.
I thought to simplify using Bernoulli inequality, but I think it doesn't works.
If I use newton expansion, instead it becomes more complicate.

Do you have any idea or tips on how to proceed?

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    I would have guessed you needed $a,b \ge 0$ to avoid issues when $a+b=0$ and $n$ is even – 2017-01-14
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    If so, note that apart from the uninformative $a=b=0$, the problem can be reduced to $c_1(1+d)\le (1+ \sqrt[\leftroot{-1}\uproot{3}n]{d})^n \le c_2(1+d)$ – 2017-01-14
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    suggesting $c_1=1$ when $d=0$ and $c_2=2^{n-1}$ when $d=1$ – 2017-01-14
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    @Henry thanks, I managed to solve by replacing a + b with d, if you want to put as an answer, I'll accept it. – 2017-01-14
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    You can put your own answer in if you wish – 2017-01-14

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