<s>[INST] Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
 [/INST] Using Cauchy-Schwarz inequality,
$$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\
\implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\
\ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\right)^2}+ \frac{-7A+24B}{2}\\
=-25 \frac{1}{2} \sqrt{A^2+B^2-4C} + \frac{-7A+24B}{2}\\
=\frac 12 \left(-25\sqrt{A^2+B^2-4C}-7A+24B\right).\blacksquare$$
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