Ta có:[tex]2x^8+\frac{3}{16}=x^8+x^8+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\geq 8\sqrt[8]{x^{16}.\frac{1}{2^{24}}}==8.x^2.\frac{1}{8}=x^2[/tex]
Dấu "=" xảy ra [tex]x^8=\frac{1}{16}\Leftrightarrow x=\pm \frac{1}{\sqrt{2}}[/tex]