Show $\prod_{i=1}^n (1 - t + t a_i) \geq 1$ when $\prod_{i=1}^n a_i = 1$

Question

Let $a_1, ..., a_n$ be elements of $\mathbb{R}^\star_+$ such that $\prod_{i=1}^n a_i = 1$.

Let $f : t \in [0, 1] \mapsto \prod_{i=1}^n (1 - t + t a_i)$. I want to show that for all $t \in [0, 1]$, we have $f(t) \geq 1$.

One can easily notice $f$ is a unitary polynomial (of degree $n$) and $f(0) = f(1) = 1$.

I have tried, but unsuccessfully, to show that $f$ is concave. I think a proof of this must strongly rely on the equality which links the $a_i$.

I also know we can write:

$$ f(t) = \sum_{I \subseteq \{ 1, ..., n \}} (1 - t)^{n - |I|} t^{|I|} \prod_{i \in I} a_i $$

But I don't know how to go further...

Any counter-example when $\prod_{i=1}^n a_i \neq 1$?

Answer 1

For one positive $a$ one has $$a^t\:=\: 1^{1-t}a^t\:\leqslant\: 1-t+ ta\tag{$\leqslant$}$$ by the weighted arithmetic-geometric mean inequality.
When applied to each factor this yields $$1\:=\:\left(\prod\nolimits_{i=1}^n a_i\right)^t \:=\: \prod_{i=1}^n a_i^t \:\leqslant\: \prod_{i=1}^n (1 - t + t a_i)\,. $$ Notice:
A geometric view on the inequality $(\leqslant)$ is that $a\mapsto a^t$ is a concave function when $\,0\leqslant t\leqslant 1$. Thus its graph lies below its tangent in $(1,1)$, given by $a\mapsto ta+1-t$.

Answer 2

Let's remark that for all $t \in [0, 1], 1 + (a_i - 1)t > 0$.

$\log(f) : t \in [0, 1] \mapsto \sum_{i=1}^n \log(1 + (a_i - 1)t)$ is subsequently well-defined and what we want to show is $\log(f) \geq 0$.

As $\log(f)(0) = \log(f)(1) = \log(1) = 0$, it is sufficient to show $\log(f)$ is concave.

To show $\log(f)$ is concave, it is sufficient to show each $t \mapsto \log(1 + (a_i - 1)t)$ is concave. The derivative of the latter function is: $t \mapsto \dfrac{a_i - 1}{1 + (a_i -1)t}$. Regardless of the sign of $a_i - 1$, one easily show this function is increasing. The sum of these functions, $\log(f)$, is still concave.