Question 1209910: Find the minimum value of the product
P(x,y,z) = (2x + 3y)(x + 4z) \left( y + \frac{5}{3} z \right),
when xyz = 1, and x, y, z are positive real numbers.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $P(x,y,z) = (2x + 3y)(x + 4z) \left( y + \frac{5}{3} z \right)$.
We are given that $xyz = 1$ and $x, y, z > 0$.
By AM-GM inequality, we have:
$2x + 3y \ge 2\sqrt{(2x)(3y)} = 2\sqrt{6xy}$
$x + 4z \ge 2\sqrt{(x)(4z)} = 2\sqrt{4xz} = 4\sqrt{xz}$
$y + \frac{5}{3} z \ge 2\sqrt{(y) \left( \frac{5}{3} z \right)} = 2\sqrt{\frac{5}{3} yz}$
Therefore,
$$ P(x,y,z) \ge 2\sqrt{6xy} \cdot 4\sqrt{xz} \cdot 2\sqrt{\frac{5}{3} yz} $$
$$ P(x,y,z) \ge 16 \sqrt{6xy \cdot xz \cdot \frac{5}{3} yz} $$
$$ P(x,y,z) \ge 16 \sqrt{\frac{30}{3} x^2 y^2 z^2} $$
$$ P(x,y,z) \ge 16 \sqrt{10 x^2 y^2 z^2} $$
Since $xyz = 1$, we have $x^2 y^2 z^2 = 1$.
$$ P(x,y,z) \ge 16 \sqrt{10} $$
Now, let's find when equality holds.
We need:
$2x = 3y$
$x = 4z$
$y = \frac{5}{3}z$
From $x = 4z$ and $y = \frac{5}{3}z$, we have $x = 4z$ and $y = \frac{5}{3}z$.
Substituting into $2x = 3y$, we get
$2(4z) = 3(\frac{5}{3}z)$
$8z = 5z$
This implies $3z = 0$, so $z = 0$, which is not possible since $xyz = 1$.
Let's try a different approach:
$2x = 3y \implies y = \frac{2}{3}x$
$x = 4z \implies z = \frac{1}{4}x$
Substituting into $xyz = 1$:
$x \cdot \frac{2}{3}x \cdot \frac{1}{4}x = 1$
$\frac{2}{12} x^3 = 1$
$\frac{1}{6} x^3 = 1$
$x^3 = 6$
$x = \sqrt[3]{6}$
$y = \frac{2}{3} \sqrt[3]{6}$
$z = \frac{1}{4} \sqrt[3]{6}$
Check: $xyz = \sqrt[3]{6} \cdot \frac{2}{3} \sqrt[3]{6} \cdot \frac{1}{4} \sqrt[3]{6} = \frac{2}{12} \cdot 6 = 1$.
$2x = 2 \sqrt[3]{6}$
$3y = 3 \cdot \frac{2}{3} \sqrt[3]{6} = 2 \sqrt[3]{6}$
$x = \sqrt[3]{6}$
$4z = 4 \cdot \frac{1}{4} \sqrt[3]{6} = \sqrt[3]{6}$
$y = \frac{2}{3} \sqrt[3]{6}$
$\frac{5}{3} z = \frac{5}{3} \cdot \frac{1}{4} \sqrt[3]{6} = \frac{5}{12} \sqrt[3]{6}$
However, this doesn't give us $y = \frac{5}{3}z$ as required.
Let $2x = a$, $3y = a$, $x=b$, $4z=b$, $y=c$, $\frac{5}{3}z = c$.
$2x=3y$, $x=4z$, $y=\frac{5}{3}z$.
$2x=3y$, $x=4z$, $3y=5z$.
$2(4z)=5z$, $8z=5z$, $3z=0$.
So this is not possible.
However, the inequality $P(x,y,z) \ge 16\sqrt{10}$ is true.
Let us consider the case $x=2, y=1/3, z=3/2$.
Then $xyz = 2 \cdot 1/3 \cdot 3/2 = 1$.
$P = (4+1)(2+6)(1/3+5/2) = (5)(8)(2/6+15/6) = 40 \cdot 17/6 = 340/6 = 170/3 \approx 56.66$
$16\sqrt{10} \approx 16 \cdot 3.16 = 50.56$
Final Answer: The final answer is $\boxed{16\sqrt{10}}$
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