Maximum Product!

Algebra Level 4

$x,y$ and $z$ are positive reals satisfying $x+y+z=1$ . If the maximum value of $x^3 y^2 z$ can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers , find $a+b$ .

The answer is 433.

1 solution

Chew-Seong Cheong
Jul 28, 2016

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

Since $x$ , $y$ and $z$ are positive, we can apply AM-GM inequality as follows:

$\begin{aligned} \frac x3 + \frac x3 + \frac x3 + \frac y2 + \frac y2 + z & \ge 6 \sqrt [6] {\frac {x^3y^2z}{3^3\cdot 2^2}} & \small \color{#3D99F6} \text{Equality occurs when } x=\frac 12, y = \frac 12, z=\frac 16 \\ \left( x + y + z\right)^6 & \ge \frac {6^6 x^3y^2z}{3^3\cdot 2^2} \\ \implies 432 x^3y^2z & \le 1 \\ x^3y^2z & \le \dfrac 1{432} \end{aligned}$

$\implies a+b = 1 + 432 = \boxed{433}$

Maximum occurs when $\frac{x}{3}=\frac{y}{2}=z=k$ . $\Rightarrow x=3k, y=2k, z=k \Rightarrow 6k = 1 \Rightarrow x=1/2, y=1/3, z=1/6$

Matt O - 4 years, 7 months ago

Thanks. I have added the line.

Chew-Seong Cheong - 4 years, 7 months ago

@Chew-Seong Cheong Sir, just a small error.......in your solution, when you have shown the equality case, I think that the LATEX code is messed up...because it is displaying words like "frac" and "small"

Aaghaz Mahajan - 2 years, 8 months ago

Maximum Product!

The answer is 433.

1 solution

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