## Problem 6.

Let $n \geq 2$ be an integer. Determine the smallest positive real number $\gamma$ such that for any positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and any real numbers $y_{1}, y_{2}, \ldots, y_{n}$ with $0 \leq$ $y_{1}, y_{2}, \ldots, y_{n} \leq \frac{1}{2}$ that satisfy $x_{1}+x_{2}+\ldots+x_{n}=y_{1}+y_{2}+\ldots+y_{n}=1$, we have

$$
x_{1} x_{2} \ldots x_{n} \leq \gamma\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)
$$
Solution.

Let $M=x_{1} x_{2} \ldots x_{n}$ and $X_{i}=\frac{M}{x_{i}}$ for $1 \leq i \leq n$. Consider the function $\varphi:(0,+\infty) \rightarrow \mathbb{R}$ defined by $\varphi(t)=\frac{M}{t}$ which is convex as can be easily verified. Since the non-negative numbers $y_{i},(1 \leq i \leq n)$, are such that $y_{1}+y_{2}+\ldots+y_{n}=1$, then applying Jensen's inequality to the function $\varphi$ we have

$$
\varphi\left(\sum_{i=1}^{n} y_{i} x_{i}\right) \leq \sum_{i=1}^{n} y_{i} \varphi\left(x_{i}\right)
$$

that is,

$$
M\left(\sum_{i=1}^{n} y_{i} x_{i}\right)^{-1} \leq \sum_{i=1}^{n} y_{i} \frac{M}{x_{i}}=\sum_{i=1}^{n} y_{i} X_{i}
$$

Now we need to find the smallest upper bound of the term on the right side of (1). Without loss of generality, we can assume that $x_{1} \leq x_{2}, \leq \ldots \leq x_{n}$ and $y_{1} \geq y_{2} \geq \ldots \geq y_{n}$. Then we have that $X_{1} \geq X_{2} \geq \ldots \geq X_{n}$ as can be easily verified. Applying the rearrangement inequality, we know that among all sums of the form $\sum_{i=1}^{n} y_{i} X_{i}$, the one that achieves the maximum value is the one obtained when $y_{1} \geq y_{2} \geq \ldots \geq y_{n}$ and $X_{1} \geq X_{2} \geq \ldots \geq X_{n}$.

Now we observe that

$$
\sum_{i=1}^{n} y_{i} X_{i}=y_{1} X_{1}+\left(y_{2} X_{2}+\ldots+y_{n} X_{n}\right) \leq y_{1} X_{1}+\left(y_{2}+\ldots+y_{n}\right) X_{2}=y_{1} X_{1}+\left(1-y_{1}\right) X_{2}
$$

Since $0 \leq y_{1} \leq 1 / 2$, we have that

$$
\begin{aligned}
\sum_{i=1}^{n} y_{i} X_{i} \leq \frac{1}{2}\left(X_{1}+X_{2}\right)=\frac{1}{2}\left(\left(x_{1}+x_{2}\right) x_{3}\right. & \left.\ldots x_{n}\right) \leq \\
& \leq \frac{1}{2}\left(\frac{\left(x_{1}+x_{2}\right)+x_{3}+\ldots+x_{n}}{n-1}\right)^{n-1}=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}
\end{aligned}
$$

where we have used the inequality between the arithmetic and geometric means and the condition $x_{1}+x_{2}+$ $\ldots+x_{n}=1$. From the above and (1), it follows that

$$
M \leq\left(\sum_{i=1}^{n} y_{i} x_{i}\right)\left(\sum_{i=1}^{n} y_{i} X_{i}\right) \leq \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\left(\sum_{i=1}^{n} y_{i} x_{i}\right)
$$

and

$$
\gamma \leq \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}
$$

If we take $x_{1}=x_{2}=\frac{1}{2(n-1)}, x_{3}=x_{4}=\ldots x_{n}=\frac{1}{n-1}$ and $y_{1}=y_{2}=\frac{1}{2}, y_{3}=y_{4}=\ldots=y_{n}=0$, then

$$
\begin{aligned}
& M=x_{1} x_{2} \ldots x_{n}=\frac{1}{4}\left(\frac{1}{n-1}\right)^{n}= \\
& \quad=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\left(y_{1} x_{1}+y_{2} x_{2}\right)=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} \sum_{i=1}^{n} y_{i} x_{i}
\end{aligned}
$$

$$
M=x_{1} x_{2} \ldots x_{n}=\frac{1}{4}\left(\frac{1}{n-1}\right)^{n}=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}\left(y_{1} x_{1}+y_{2} x_{2}\right)=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} \sum_{i=1}^{n} y_{i} x_{i}
$$

and we conclude that

$$
\gamma=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}
$$