Prove the following inequality:
$$\sum_{i = 1}^n a_i – \sum_{i = 1}^n \frac{1}{a_i} \geq n\left(\sqrt[n]{\prod_{i = 1}^na_i}- \frac{1}{\sqrt[n]{\prod_{i = 1}^na_i}}\right)$$
Where all the variables are greater or equal than $1$.
This inequality is a generalization of a standard one (actually the base case in the induction method I'm trying to use, as presented later on the post). I am thinking at the special type of induction called Cauchy Induction. Here is my progress:
Base case. It's a well known that, for $x, y \geq 1$:
$$x + y – \frac{1}{x} – \frac{1}{y} \geq \sqrt{xy} – \frac{1}{\sqrt{xy}}$$
In fact, we can rewrite the form as:
$$(xy – 1)\left(\sqrt{x} – \sqrt{y}\right)^2$$
Inductive step 1 Prove that $p(n) \to p(2n)$. Let's consider $a_1, a_2, \dots, a_n$, $b_1, b_2, \dots, b_n$ $2n$ numbers. From $p(n)$, we have that:
$$LHS \geq n\left(\sqrt[n]{\prod_{i = 1}^na_i}- \frac{1}{\sqrt[n]{\prod_{i = 1}^na_i}}\right) + n\left(\sqrt[n]{\prod_{i = 1}^nb_i}- \frac{1}{\sqrt[n]{\prod_{i = 1}^nb_i}}\right)$$
And, by using the base case:
$$LHS \geq 2n \left(\sqrt[2n]{\prod_{i = 1}^na_ib_i}- \frac{1}{\sqrt[2n]{\prod_{i = 1}^na_ib_i}}\right)$$
However, I could not prove backwards induction step. Any hint will be appreciated.
Best Answer
A proof without induction.
Let $a_i=e^{x_i}$ and $f(x)=e^x-\frac{1}{e^x}.$
Thus, since $x_i\geq0$ and $f''(x)=e^x-\frac{1}{e^x}\geq0$ for $x\geq0$,
we obtain that $f$ is a convex function on $[0,+\infty).$
Thus, by Jensen we obtain: $$\sum_{i=1}^nf(x_i)\geq nf\left(\frac{\sum\limits_{i=1}^nx_i}{n}\right),$$ which is your inequality