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Let $x$ and $y$ are variables satisfying $3x-4y\propto\sqrt{xy}$. Then show that $x^2+y^2\propto xy$, where $y∝x$ implies $y$ directly proportional to $x$. My attempt... $3x-4y=k\sqrt{xy}$ for some real $k$. This implies $(3x-4y)^2=k^2xy$ implies $9x^2+16y^2=(k^2+24)xy$ implies $x^2+y^2=((k^2+24)x-7y)y/9$. So i have to assert now that $[(k^2+24)x-7y]\propto x$. But I cant proceed from here. Please help me to solve this.

abcdmath
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  • Just an aside question: I've been seeing that infinity-ish symbol a lot lately. Would someone care to explain what does it mean? I guess it is some sort of asymptotic relation (?) @vadim123 Lol, it makes me feel better to know that I am not the only one. ;) – Jonatan B. Bastos Dec 11 '17 at 17:03
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    $y\propto x$ means that $y$ is directly proportional to $x$. – dromastyx Dec 11 '17 at 17:04
  • $x$ and $y$ are specific real numbers? In that case everything is proportional to everything. There are no variables....?? – John Brevik Dec 11 '17 at 17:17
  • No here $x$ and $y$ are variables, but are belonging to reals. @John Brevik – abcdmath Dec 11 '17 at 17:21

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Let $t = \sqrt{\frac{x}{y}}$. Observe that $3t - \frac{4}{t} = k$, or $3t^2 -kt - 4 = 0$, giving $t = \frac{k \pm \sqrt{k^2 + 12}}{6}$, that is $t$ is constant.

On the other hand, $\frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x} = t^2 + t^{-2}$ is also constant, and the claim follows.

user58697
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