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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

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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.

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From where you have left of,

For real $x-2y,$ $$(x-2y)^2\ge0\implies x^2+4y^2\ge4xy$$

Alternatively, using AM-GM inequality, $$\dfrac{x^2+4y^2}2\ge\sqrt{4x^2y^2}=?$$

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By C-S $$(2x+3y)\left(\frac{8}{x}+\frac{3}{y}\right)\geq\left(\sqrt{2x\cdot\frac{8}{x}}+\sqrt{3y\cdot\frac{3}{y}}\right)^2=49.$$ The equality occurs for $x=2$ and $y=1$, which says that we got a minimal value.

Done!

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