2. CMR : $\frac{-1}{2} \leq \frac{(a + b)(1-ab)}{(a^2 + 1)(b^2 + 1)} \leq \frac{1}{2}$
Lời giải.
Bất đẳng thức cần chứng minh tương đương với:
$$\left | \dfrac{\left ( a+b \right )\left ( 1-ab \right )}{\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )} \right |\leq \dfrac{1}{2}$$
Đặt $a=\tan \alpha$, $b=\tan \beta$ khi đó:
\begin{align*} \left | \dfrac{\left ( a+b \right )\left ( 1-ab \right )}{\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )} \right | &=\left | \dfrac{\left ( \tan \alpha +\tan \beta \right )\left ( 1-\tan \alpha \tan \beta \right )}{\left ( \tan ^{2}\alpha +1 \right )\left ( \tan ^{2}\beta +1 \right )} \right | \\ &=\left | \cos ^{2}\alpha \cos ^{2}\beta \dfrac{\sin \left ( \alpha +\beta \right )\left ( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right )}{\cos ^{2}\alpha \cos ^{2}\beta } \right | \\ &=\left | \sin \left ( \alpha +\beta \right )\cos \left ( \alpha +\beta \right ) \right | \\ &=\dfrac{1}{2}\left | \sin 2\left ( \alpha +\beta \right ) \right |\leq \dfrac{1}{2} \end{align*}
Hoặc có thể làm cách khác như sau:
Lời giải.
Ta có:
$$\dfrac{\left ( a+b \right )\left ( 1-ab \right )}{\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )}+\dfrac{1}{2}=\dfrac{\left ( ab-a-b-1 \right )^{2}}{2\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )}\geq 0$$
$$\dfrac{1}{2}-\dfrac{\left ( a+b \right )\left ( 1-ab \right )}{\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )}=\dfrac{\left ( ab+a+b-1 \right )^{2}}{2\left ( a^{2}+1 \right )\left ( b^{2}+1 \right )}\geq 0$$