Question

Sean $x_1,$ $x_2,$ $\ldots,$ $x_6$ números reales no negativos tales que $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 =$ $1,$ y $x_1x_3x_5 + x_2x_4x_6 \ge \frac{1}{540}.$ Sean $p$ y $q$ enteros positivos coprimos tales que $\frac{p}{q}$ es el valor máximo posible de $\begin{aligned} &x_1x_2x_3 + x_2x_3x_4 \\ &\quad {}+ x_3x_4x_5 + x_4x_5x_6 \\ &\quad {}+ x_5x_6x_1 + x_6x_1x_2. \end{aligned}$ Halla $p + q.$

Let $x_1,$ $x_2,$ $\ldots,$ $x_6$ be nonnegative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 =$ $1,$ and $x_1x_3x_5 + x_2x_4x_6 \ge \frac{1}{540}.$ Let $p$ and $q$ be positive relatively prime integers such that $\frac{p}{q}$ is the maximum possible value of $\begin{aligned} &x_1x_2x_3 + x_2x_3x_4 \\ &\quad {}+ x_3x_4x_5 + x_4x_5x_6 \\ &\quad {}+ x_5x_6x_1 + x_6x_1x_2. \end{aligned}$ Find $p + q.$

Respuesta

Solución:

Sea $r = x_1x_3x_5 + x_2x_4x_6$ y sea $s$ la suma cíclica en cuestión. Al expandir $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ se producen ocho productos triples, que son exactamente los seis términos de $s$ junto con los dos términos de $r.$ Así que $r + s =$ $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6),$ y por AM-GM esto es a lo sumo $\left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

Por lo tanto $\begin{aligned} s &\le \frac{1}{27} - r \\ &\le \frac{1}{27} - \frac{1}{540} \\ &= \frac{20 - 1}{540} = \frac{19}{540}. \end{aligned}$ La igualdad requiere $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =$ $\frac{1}{3}$ con $r = \frac{1}{540}:$ toma $x_1 = x_3 = \frac{3}{10},$ $x_5 = \frac{1}{60},$ $x_2 = \frac{19}{60},$ $x_4 = x_6 = \frac{1}{30}.$ Entonces $r = \frac{9}{6000} + \frac{19}{54000} = \frac{100}{54000} = \frac{1}{540},$ como se requería.

Así que el máximo es $\frac{19}{540},$ y $p + q = 19 + 540 = 559.$

Let $r = x_1x_3x_5 + x_2x_4x_6$ and let $s$ be the cyclic sum in question. Expanding $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ produces eight triple products, which are exactly the six terms of $s$ together with the two terms of $r.$ So $r + s =$ $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6),$ and by AM-GM this is at most $\left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

Therefore $\begin{aligned} s &\le \frac{1}{27} - r \\ &\le \frac{1}{27} - \frac{1}{540} \\ &= \frac{20 - 1}{540} = \frac{19}{540}. \end{aligned}$ Equality needs $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =$ $\frac{1}{3}$ with $r = \frac{1}{540}:$ take $x_1 = x_3 = \frac{3}{10},$ $x_5 = \frac{1}{60},$ $x_2 = \frac{19}{60},$ $x_4 = x_6 = \frac{1}{30}.$ Then $r = \frac{9}{6000} + \frac{19}{54000} = \frac{100}{54000} = \frac{1}{540},$ as required.

So the maximum is $\frac{19}{540},$ and $p + q = 19 + 540 = 559.$

2011 AIME II Problema 9

Solución:

El Problema 9 en otros años