Chứng minh: \(-2\sqrt{2}-2\leq \frac{x^2-(x-4y)^2}{x^2+4y^2}\leq 2\sqrt{2}-2\)

Nếu y = 0 khi đó \(\small x\neq 0\)  ta có
\(\small \frac{x^2-(x-4y)^2}{x^2+4y^2}=0\) bất đẳng thức hiển nhiên đúng
Nếu \(\small y\neq 0\) khi đó \(\small \Leftrightarrow -2\sqrt{2}-2\leq \frac{x^2-(x-4y)^2}{x^2+4y^2}\leq 2\sqrt{2}-2\)
\(\small \Leftrightarrow -2\sqrt{2}-2\leq \frac{(\frac{x}{2y})^2-(\frac{x}{2y}-2)^2}{(\frac{x}{2y})^2+1}\leq 2\sqrt{2}-2 \ \ (1)\)
Đặt \(\small \frac{x}{2y}=tant\), khi đó \(\small \Leftrightarrow -2\sqrt{2}-2\leq \frac{tan^2t-(tant-2)^2}{tan^2t+1}\leq 2\sqrt{2}-2\)
\(\small (1)\Leftrightarrow -2\sqrt{2}-2\leq cos^2t(4tant-4)\leq 2\sqrt{2}-2\)
\(\small \Leftrightarrow -2\sqrt{2}-2\leq 2sin2t-4cos^2t\leq 2\sqrt{2}-2\)
\(\small \Leftrightarrow -2\sqrt{2}-1\leq sin2t-2cos^2t\leq \sqrt{2}-1\)
\(\small \Leftrightarrow -\sqrt{2}\leq sin2t+1-2cos^2t\leq \sqrt{2}\)
\(\small \Leftrightarrow -\sqrt{2}\leq sin2t-cos2t\leq \sqrt{2}\)
\(\small \Leftrightarrow -\sqrt{2}\leq \sqrt{2}sin(2t-\frac{\pi}{4})\leq \sqrt{2}\)
\(\small \Leftrightarrow -1\leq sin(2t-\frac{\pi}{4})\leq 1 \ (2)\)
Vì (2) đúng suy ra đpcm.