Let $n\geq 2$ an integer. Find the least value of $\gamma$ such that for any positive real numbers $x_1,x_2,...,x_n$ with $x_1+x_2+...+x_n=1$ and any real $y_1+y_2+...+y_n=1$ and $0\leq y_1,y_2,...,y_n\leq \frac{1}{2}$ the following inequality holds: $$x_1x_2...x_n\leq \gamma \left(x_1y_1+x_2y_2+...+x_ny_n\right)$$
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Tags: national olympiad, inequalities, Jensen, rearrangement inequality, Spain, ESP
ESCHen99
17.05.2016 12:36
Any solution?
socrates
16.11.2016 00:07
Up! Up!
Mprog.
12.07.2018 13:37
????????????
seoneo
12.07.2018 14:50
The problem is to maximize \[ \lambda = \frac{x_1 x_2 \ldots x_n}{x_1 y_1 + x_2 y_2 + \cdots + x_n y_n } \]Without loss of generality suppose that $x_1 \ge x_2 \ge x_3 \ge \cdots \ge x_n$. Tentatively, freeze $x_j$s. To maximize \[ \lambda = \frac{x_1 x_2 \ldots x_n}{x_1 y_1 + x_2 y_2 + \cdots + x_n y_n } \]is to minimize \[ x_1 y_1 + x_2 y_2 + \cdots + x_n y_n \]From the linearity, we get the minimum \[ \frac{1}{2} \left( x_{n-1} + x_n \right) \]so in this case \[ \lambda \le \frac{2 x_1 x_2 \ldots x_n}{x_{n-1} + x_n} \le x_1 x_2 \cdots x_{n-2} \sqrt{x_{n-1}x_n} \]Let $a$ be the arithmetic mean of $x_1, x_2 , \ldots , x_{n-2}$ and $b$ be the arithmetic mean of $x_{n-1}$ and $x_n$. Clearly $a \ge b$ and $(n-2)a+2b =1$. \[ x_1 x_2 \cdots x_{n-2} \sqrt{x_{n-1}x_n} \le a^{n-2}b \le\frac{1}{2} \left( \frac{a \times (n-2) + 2b } {n-1}\right)^{n-1} = \frac{1}{2(n-1)^{n-1}} \] This equality is really achievable when $x_1 = x_2 = \cdots = x_{n-2} =a$, $x_{n-1}=x_n=b$, $b=2a=\frac{1}{n-1}$.