Question

Cho tam giác ABC có: tanA+tanC=2tanB. CMR: cosA+cosC\(\le\dfrac{3\sqrt{2}}{4}\)

Answer 1

\(\Rightarrow \tan A+\tan C=2\tan B\)

\(\Leftrightarrow \frac{\sin\left ( A+C \right )}{\cos A\cos C}=2\cdot\frac{\sin\left ( A+C \right )}{\cos B}\\\)

\(\Rightarrow \cos B=2\cos A\cos C\)

\(\Leftrightarrow 2\cos B=\cos(A-C)\)

\(\left (\cos A+\cos C \right )^2=\cos^2 A+\cos^2 C+2\cos A\cos C\\=\frac{\cos2A+\cos2C}{2}+1+\cos B\\=-\cos(B)\cos(A-C)+1+\cos B \\=-2\cos^2B+\cos B+1 \le \frac{9}{8}\\\Rightarrow \cos A+\cos C\le \frac{3\sqrt2}{4}\)

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