<s>[INST] Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$
But I cannot get the minimum value.
 [/INST] You are almost there: Note that by Cauchy-Schwarz you  did
$$25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\,.$$
Thus
$$\left|\frac{8x-3y}{\sqrt{4x^2+y^2}}\right|\le 5\tag {*}\\[5ex]
\implies -5\le \frac{8x-3y}{\sqrt{4x^2+y^2}}\le 5$$
The important part was in step (*). The modulus was crucial as $8x-3y$ could be negative too.
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