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.
Best Answer
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.