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Show that $$1+(x_1x_2...x_n)^{\frac{1}{n}} \leq [(1+x_1)(1+x_2)...(1+x_n)]^{\frac{1}{n}}, \forall x_i \geq 0, \ i = 1,2,3...,n$$

So, I have to make this function something like this:

$$f(t_1x_1+t_2x_2+...t_nx_n) \leq t_1f(x_1)+t_2f(x_2)...+t_nf(x_n)$$ Also, I need to choose an $f(x)$ such that the inequality holds and $f''(x) > 0$ to show that this is a convex.

Ok, so it is getting really difficult for me as I don't understand what to choose as $t_i$ and what to choose as $f(x)$. All I can do is this:

$$1+[e^{\log(x_1x_2...x_3)^{\frac{1}{n}}}] \leq e^{\log[(1+x_1)(1+x_2)...(1+x_n)]^{\frac{1}{n}}}$$

Maybe I can choose $f(x) = \log(x)$, but that'd give me $f''(x) < 0.$ Can anyone please help me figure out what $t_i$ and $f(x)$ should be in order to make it a convex? Thanks.

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  • $\begingroup$ Maybe you could use induction on $n$ and then just have to prove the case for $n=2$. $\endgroup$ – bgins Jul 24 '15 at 17:55
  • $\begingroup$ @bgins, I think I could. But, I am trying to learn it in the way as above. $\endgroup$ – Jellyfish Jul 24 '15 at 18:01
  • $\begingroup$ Maybe try $f(x)=\log(1+x)$, $t_i=1$ and deal with $f$ being concave in stead of convex (so $-f$ would be convex). $\endgroup$ – bgins Jul 24 '15 at 18:12
  • $\begingroup$ Related: math.stackexchange.com/questions/29357/… $\endgroup$ – user940 Jul 24 '15 at 21:00
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You may take $g(t)=\log(1+e^t)$. We have: $$ g''(t) = \frac{e^t}{(1+e^t)^2} $$ hence $g$ is a convex function. Let $x_i=e^{t_i}$. Jensen's inequality hence gives:

$$ \log\left(1+\exp\left(\frac{t_1+\ldots+t_n}{n}\right)\right)\leq \frac{1}{n}\sum_{i=1}^{n}\log(1+e^{t_i})$$ and by exponentiating the previous line: $$ 1+\left(x_1\cdot\ldots\cdot x_n\right)^{\frac{1}{n}}\leq\left((1+x_1)\cdot\ldots\cdot(1+x_n)\right)^{\frac{1}{n}}$$ i.e. the super-additivity of the geometric mean, follows.

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You can invoke Maclaurin's Inequality (https://en.wikipedia.org/wiki/Maclaurin%27s_inequality). Let $S_k$ be the average symmetric sum of degree $k$ for $x_1,x_2,\ldots,x_n$ and $k=0,1,2,\ldots,n$. Then, $\sqrt[k]{S_k}\geq \sqrt[n]{S_n}$ for every $k$. Note also that $1=S_0=S_1^0$ (where we take $0^0$ to be $1$), and $S_1\geq\sqrt[k]{S_k}$ for all positive $k$. Now, since $\prod_{i=0}^n\,\left(1+x_i\right)=\sum_{k=0}^n\,\binom{n}{k}\,S_k$, we get $$\prod_{i=0}^n\,\left(1+x_i\right) \geq \sum_{k=0}^n\,\binom{n}{k}\,\left(\sqrt[n]{S_n}\right)^k=\left(1+\sqrt[n]{S_n}\right)^n\,.$$ Also, you can show that $$\prod_{i=0}^n\,\left(1+x_i\right) \leq \sum_{k=0}^n\,\binom{n}{k}\,S_1^k=\left(1+S_1\right)^n\,.$$ Hence, $$1+\sqrt[n]{\prod_{i=1}^n\,x_i}\leq\sqrt[n]{\prod_{i=1}^n\,\left(1+x_i\right)}\leq 1+\frac{\sum_{i=1}^n\,x_i}{n}\,.$$ However, you can also achieve the inequalities above by applying AM-GM too.

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