Bài 1: CMR $\frac{-1}{2} \leq \frac{(a+b)(1-ab)}{(1+a^2)(1+b^2)} \leq \frac{1}{2}$
C1:$\frac{(a+b).(1-ab)}{(1+a^{2}).(1+b^{2})}+\frac{1}{2}=\frac{(ab-a-b-1)^2}{2(1+a^2)(1+b^2)} \geq 0$
$\frac{1}{2}-\frac{(a+b).(1-ab)}{(1+a^{2}).(1+b^{2})} =\frac{(ab+a+b-1)^2}{2(1+a^2)(1+b^2)} \geq 0$
C2:Ta có:
Áp dụng $ab\leq \frac{(a+b)^2}{4}$ ta có:
$(a^2+b^2+2ab)(a^2b^2-2ab+1)\leq \frac{(a^2+b^2+a^2b^2+1)^2}{4}=\frac{(a^2+1)^2(b^2+1)^2}{4}\Rightarrow \frac{(a+b)^2(ab-1)^2}{(a^2+1)^2(b^2+1)^2}\leq \frac{1}{4}\Rightarrow Q.E.D$
C3: Bất đẳng thức cần chứng minh tương đương với:
\[\left| {\frac{{\left( {a + b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}}} \right| \le \frac{1}{2}\]
Đặt $a = \tan \alpha ,b = \tan \beta $. Khi đó: \[\left| {\frac{{\left( {a + b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}}} \right| = \left| {\frac{{\left( {\tan \alpha + \tan \beta } \right)\left( {1 - \tan \alpha \tan \beta } \right)}}{{\left( {1 + {{\tan }^2}\alpha } \right)\left( {1 + {{\tan }^2}\beta } \right)}}} \right|\]
\[ = \left| {{{\cos }^2}\alpha {{\cos }^2}\beta \frac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \alpha \cos \beta }}.\frac{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}} \right|\]
\[ = \left| {\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right)} \right| = \frac{1}{2}\left| {\sin 2\left( {\alpha + \beta } \right)} \right| \le \frac{1}{2}\,\,\left( \text{đpcm} \right)\]