$cos^{2} A+cos^{2}B+cos^{2}C\geq \frac{3}{4 }$
Ta có: $\cos^{2} A+\cos^{2}B+\cos^{2}C$$={{\cos }^{2}}A+\frac{1+\cos 2B}{2}+\frac{1+\cos 2C}{2}$
$={{\cos }^{2}}A-\cos \left( B-C \right).\cos A+1$$={{\left[ \cos A-\frac{1}{2}\cos \left( B-C \right) \right]}^{2}}+\frac{1}{4}{{\sin }^{2}}\left( B-C \right)+\frac{3}{4}\ge \frac{3}{4}$
Dấu “=” xảy ra khi $A=B=C={{60}^{0}}$.
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