I have a tricky inequality (related to some previous ones that I posted) which I am yet again stuck trying to solve. I have confirmed that it is true through simulation (at least up until overflow errors occur).
Let $x_j$ be a natural number greater than equal to 2, for all $j$. Define also $\bar{x} = \frac{1}{N} \sum_{j=1}^N x_j$ and $\dot{x} = \prod_{j=1}^N x_j$. I would like to show that \begin{align} N\left(1 + \bar{x}\right)^2 \left(1 + \frac{2}{\bar{x}}\right) \leq \left(1 + \dot{x}\right)^2 \left(1 + \frac{2}{\dot{x}}\right) \; ; \end{align} or, equivalently, \begin{align} N\left(5 + 4 \bar{x} + \frac{2}{\bar{x}} + \bar{x}^2\right) \leq \left(5 + 4\dot{x} + \frac{2}{\dot{x}} + \dot{x}^2 \right) \; . \end{align}
I'm really not sure the best approach for this. Would induction be a good approach or should a more direct approach be taken?
Any help at all would be appreciated.
It is straight-forward to verify that the function $$ f(x) = (1+x)^2 \left( 1 + \frac 2x\right) $$ is increasing for $x \ge 2$ and also satisfies $f(2x) \ge 2f(x)$ for $x \ge 2$.
We have $\dot{x} = \prod_{j=1}^N x_j \ge 2^{N-1} \bar x$ since all factors are $\ge 2$ and at least one factor is $\ge \bar x$. It follows that $$ f(\dot x) \ge f(2^{N-1} \bar x) \ge 2^{N-1} f(\bar x) \ge N f(\bar x) \, , $$ using Bernoulli's inequality in the last step: $$ 2^{N-1} = (1+1)^{N-1} \ge 1+(N-1) = N \, . $$