Minimum Value of expression 2 What is the minimum value of $$(2x+3y)(8/x + 3/y)$$ when $x$ and $y$ are positive?
I have tried expanding the equation but I did not get anything
I got :  $$25 + \frac{6(4y^2 +x^2)}{xy}$$ but I don't know how to answer after this.
I have also tried AM-GM but I couldn't manipulate the variables to cancel  out and get the inequality.
This is from Sipnayan 2017

You were on the right track:
$$\begin{align}(2x+3y)\left(\frac{8}{x} + \frac{3}{y}\right) &= 25+24 \frac{y}{x}+6\frac{x}{y}\\
&=25+12\left(\frac{2y}{x}+\frac{x}{2y}\right) \\
&\geq 25+12 \times 2 \\
&=49\end{align}$$
and notice that $x=2,y=1$ achieves this bound, so the minimum value is $49$.
The bound $\frac{2y}{x} + \frac{x}{2y} \geq 2$ can be proved from AM-GM, or from letting $t = \frac{2y}{x}$ and considering the function $f(t) = t + \frac{1}{t}$ and using calculus.
