<s>[INST] If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$. I am stuck on the following problem which one of my friends gave me: 

If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$. 

The problem could have been easier if I had to prove $4x^2+9y^2 \propto xy$ but this problem has stumped me. 
Can someone please help ? Thanks in advance for your time.
 [/INST] $$\text{A) }2x+3y \propto \sqrt{xy}\qquad \text{B) }9x^2+4y^2 \propto xy$$
From A we get
$$4 x^2+12 x y+9 y^2 \propto xy$$
Combining A and B we get
$$4 x^2+12 x y+9 y^2 = k(9x^2+4y^2)$$
$$\Rightarrow 4 x^2+12 x y+9 y^2 = 9kx^2+4ky^2$$
From this we see that no $k$ will give an $xy$ term, so the statement is false. If we continue, we get
$$ 12xy = x^2(9k-4)+y^2(4k-9)$$
$$ xy = c_1x^2+c_2y^2$$
This is even more nonsensical. Further, if we substitute in the definition of $k$ from A we get
$$4 x^2+12 x y+9 y^2 = \frac{2x+3y}{\sqrt{xy}}(9x^2+4y^2)$$
Solving this for $y$, we get
$$y = \frac{-2x}{3}$$
Since the signs of $x$ and $y$ differ for all non-zero $x$,$y$ we conclude that the original statement only holds for $x$,$y = 0$, and does not hold in general
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