IF $x , y , z$ are arbitary positive real numbers satisfying the equation
$$ 4xy + 6yz + 8xz = 9$$
Find the maximum value of the product $xyz$.
I dont know from where to begin .
3 variables and one equation.
How I can achieve this?
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Sign up to join this communityIF $x , y , z$ are arbitary positive real numbers satisfying the equation
$$ 4xy + 6yz + 8xz = 9$$
Find the maximum value of the product $xyz$.
I dont know from where to begin .
3 variables and one equation.
How I can achieve this?
Arithmetic mean is $\ge$ Geometric mean, i.e. $${{4xy + 6yz + 8xz}\over3} \ge {{(4xy\cdot 6yz\cdot 8xz)}}^{1/3}.$$
If we want to use the AGM inequality without loss we have to "symmetrize" the three variables. Therefore we replace $x$, $y$, $z$ by $$x':=4x,\quad y':=3y,\quad z':=6z\ .$$ The AGM inequality then says that $$\root 3\of {x'^2y'^2z'^2}\leq {x'y' + y'z' + z' x'\over 3}=4xy + 6 yz+8 zx=9\ ,$$ i.e., $x'y'z'\leq 27$, with equality iff $x'=y'=z'=3$. It follows that $$xyz = {x'y'z'\over 72}\leq{3\over 8}$$ with equality iff $x={3\over4}$, $y=1$, $z={1\over2}$.
You can also do it by using weighted arithmetic mean and geometric mean inequality,
$$\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n} > [{x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}}]^{\frac{1}{m_1m_1m_2m_3.....m_n}}$$