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Áp dụng BĐT Cauchy ta có:$\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le\dfrac{1}{2}\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)$Tương tự suy ra:$\sum_{A,B,C}\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le2(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2})$Ta chỉ cần chứng minh: $\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\le\cot A+\cot B+\cot C$Ta có:$\cot A+\cot B=\dfrac{\sin(A+B)}{\sin A\sin B}=\dfrac{2\sin C}{\cos(A-B)-\cos(A+B)}\ge\dfrac{2\sin C}{1+\cos C}=\tan\dfrac{C}{2}$Tương tự: $\cot B+\cot C\ge\tan\dfrac{A}{2},\cot C+\cot A\ge\tan\dfrac{B}{2}$Cộng 3 BĐT trên ta có đpcm.

Áp dụng BĐT Cauchy ta có:$\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le\dfrac{1}{2}\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)$Tương tự suy ra:$\sum_{A,B,C}\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le2(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2})$Ta chỉ cần chứng minh: $\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\le\cot A+\cot B+\cot C$Ta có:$\cot A+\cot B=\dfrac{\sin(A+B)}{\sin A\sin B}=\dfrac{2\sin C}{\cos(A-B)-\cos(A+B)}\ge\dfrac{2\sin C}{1+\cos C}=2\tan\dfrac{C}{2}$Tương tự: $\cot B+\cot C\ge2\tan\dfrac{A}{2},\cot C+\cot A\ge2\tan\dfrac{B}{2}$Cộng 3 BĐT trên ta có đpcm.