ta có [tex]1+16a^{4}\geq 2\sqrt{16x^{4}}=8x^{2}[/tex](am-gm)
[tex]\rightarrow \frac{a^{2}}{1+16a^{4}}\leq \frac{a^{2}}{8a^{2}}=1/8[/tex]
ttuong tu [tex]\frac{b^{2}}{1+16b^{4}}\leq 1/4[/tex]
cong ve vs cve các bdt tren ta dc vt<=1/4
dau = xay ra khi a=b=1/2
ta có [tex]1+16a^{4}\geq 2\sqrt{16x^{4}}=8x^{2}[/tex](am-gm)
[tex]\rightarrow \frac{a^{2}}{1+16a^{4}}\leq \frac{a^{2}}{8a^{2}}=1/8[/tex] ttuong tu $\frac{b^{2}}{1+16b^{4}}\leq 1/4$
cong ve vs cve các bdt tren ta dc vt<=1/4
dau = xay ra khi a=b=1/2