<s>[INST] Find the minimum value of $3x + 4y$ The minimum value of  $3x + 4y$  subject to the condition 


$$x^2 y^3  = 6$$ 


and $x$ and $y$ are positive .
 [/INST] $$3x+4y = 2(\frac{3x}{2})+3(\frac{4y}{3})$$
According to weighted arithmetic mean and weighted geometric mean inequality,

$$\left(\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n}\right)^{m_1m_1m_2m_3.....m_n} \ge x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}$$

$$\frac{2(\frac{3x}{2})+3(\frac{4y}{3})}{2+3} \ge[[\frac{3x}{2}]^2[\frac{4y}{3}]^3]^\frac{1}{2+3}$$
$$\frac{3x+4y}{5} \ge[\frac{16x^2y^3}{3}]^\frac{1}{5}$$
$x^2y^3=6$
$$\frac{3x+4y}{5} \ge{32}^{1/5}$$
$$3x+4y \ge 10$$
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