Problem 4. Find the smallest positive real number $\lambda$ such that for any real numbers $a_{1}, a_{2}, a_{3} \in\left[0, \frac{1}{2}\right]$ and $b_{1}, b_{2}, b_{3} \in(0, \infty)$ with $\sum_{i=1}^{3} a_{i}=\sum_{i=1}^{3} b_{i}=1$, we have

$$
b_{1} b_{2} b_{3} \leq \lambda\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right)
$$
Solution. Let the function $f:(0, \infty) \rightarrow \mathbb{R}, f(x)=\frac{b_{1} b_{2} b_{3}}{x}$. The function $f$ is convex $\ldots .1 \mathrm{p}$

We apply Jensen's inequality: $\frac{b_{1} b_{2} b_{3}}{\sum_{i=1}^{3} a_{i} b_{i}}=f\left(\sum_{i=1}^{3} a_{i} b_{i}\right) \leq \sum_{i=1}^{3} a_{i} f\left(b_{i}\right) \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \mathrm{m}$

Assume that $b_{1} \leq b_{2} \leq b_{3}$. Then $f\left(b_{1}\right) \geq f\left(b_{2}\right) \geq f\left(b_{3}\right)$.

We have

$$
\begin{aligned}
& \frac{b_{1} b_{2} b_{3}}{\sum_{i=1}^{3} a_{i} b_{i}} \leq a_{1} f\left(b_{1}\right)+a_{2} f\left(b_{2}\right)+a_{3} f\left(b_{3}\right) \\
& \leq a_{1} f\left(b_{1}\right)+\left(a_{2}+a_{3}\right) f\left(b_{2}\right) \\
& =a_{1} f\left(b_{1}\right)+\left(1-a_{1}\right) f\left(b_{2}\right) \\
& =f\left(b_{2}\right)+a_{1}\left(f\left(b_{1}\right)-f\left(b_{2}\right)\right) \\
& \leq f\left(b_{2}\right)+\frac{1}{2}\left(f\left(b_{1}\right)-f\left(b_{2}\right)\right) \\
& =\frac{1}{2}\left(b_{1}+b_{2}\right) b_{3} \\
& \leq \frac{1}{2}\left(\frac{1}{2}\right)^{2}=\frac{1}{8}
\end{aligned}
$$

In the last inequality, we used the inequality of means and the condition from the statement. . ........... 3 p

It is easily verified that for $a_{1}=a_{2}=\frac{1}{2}, a_{3}=0$ and $b_{1}=b_{2}=\frac{1}{4}, b_{2}=\frac{1}{2}$, the equality is achieved

![](https://cdn.mathpix.com/cropped/2024_06_07_306fdcac2e80fe22fd4fg-3.jpg?height=54&width=1589&top_left_y=442&top_left_x=230)

![](https://cdn.mathpix.com/cropped/2024_06_07_306fdcac2e80fe22fd4fg-3.jpg?height=56&width=1526&top_left_y=487&top_left_x=294)