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Ta có $(\sin^2a+\sin^2b)^2 =\sin^4a+\sin^4b+2\sin^2a\sin^2b\ge \sin^4a+\sin^4b$ $\Leftrightarrow \sin^2a+\sin^2b \ge \sqrt{\sin^4a+\sin^4b}$ Áp dụng BĐT dạng $\sqrt{a^2+b^2}+\sqrt{c^2+d^2} \ge \sqrt{(a+c)^2+(b+d)^2}$ ta được $\sqrt{ \cos^4a+\cos^4b}+\sin^2a+\sin^2b \geq \sqrt{ \cos^4a+\cos^4b}+\sqrt{\sin^4a+\sin^4b}\geq\sqrt{(\sin^2a+\cos^2a)^2+(\sin^2b+\cos^2b)^2}= \sqrt{ 2}$
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