Chứng minh: [tex]\frac{1}{1 + Cos2a} + \frac{1}{1 + cos4a} + \frac{1}{1 - Cos6a} > 2[/tex]
Ta có [tex]VT\geq \frac{9}{3+cos2a+cos4a-cos6a}=\frac{9}{3+2cosacos3a-2cos^{2}3a+1}[/tex]=[tex]\frac{9}{3-2(cos3a-\frac{1}{2}cosa)^{2}+\frac{1}{2}cos^{2}a}> \frac{9}{3+1,5}=2[/tex]..