Assume that for $a,b>0$ and any $0 < t< 1$

$$a^tb^{1-t} ≤ ta+(1-t)b$$

Prove given $a_1,a_2,...,a_n ≥ 0$, $b_1,b_2,...,b_n \geq 0$ and $b_1+b_2+...+b_n=1$

We have

$$\left(\sum_{i = 1}^n\frac{b_i}{a_i}\right)^{-1} \leq \prod_{i = 1}^n a_i^{b_i} \leq \sum_{i = 1}^n b_i*a_i$$

Let $c_i = \dfrac{1}{a_i}$, then using weighted AM-GM inequality we have:
$\displaystyle \sum_{k = 1}^n \dfrac{b_k}{a_k} = \displaystyle \sum_{k = 1}^n b_k\cdot c_k \geq \displaystyle \prod_{k=1}^n c_k^{b_k}$. So:
$\displaystyle \sum_{k=1}^n \dfrac{b_k}{a_k}\cdot \displaystyle \prod_{k=1}^n a_k^{b_k} \geq \displaystyle \prod_{k=1}^n (c_k\cdot a_k)^{b_k} = 1$ since $c_k\cdot a_k = 1$ for all $k$'s.
The right inequality is proven by induction on $n$.