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Ta có :
$\cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}
=\dfrac{1}{2}\cos^2\dfrac{A}{2}\left[\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2}\right]$
Vì thế :
$\cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\le\dfrac{1}{2}\cos^2\dfrac{A}{2}(1+\sin\dfrac{A}{2})$
$\Leftrightarrow
\cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\le\dfrac{1}{4}(1+\sin\dfrac{A}{2})(1+\sin\dfrac{A}{2})(2-2\sin\dfrac{A}{2})$
Áp dụng bất đẳng thức Cauchy, ta có :
$(1+\sin\dfrac{A}{2})(1+\sin\dfrac{A}{2})(2-2\sin\dfrac{A}{2})\le\left[\dfrac{(1+\sin\dfrac{A}{2})+(1+\sin\dfrac{A}{2}) + (2-2\sin\dfrac{A}{2})}{3} \right]^3$
$ \Rightarrow (1+\sin\dfrac{A}{2})^2(2-2\sin\dfrac{A}{2}) \le \dfrac{64}{27}$
Suy ra:
$\cos\dfrac{A}{2}\sqrt{\cos\dfrac{B}{2}\cos\dfrac{C}{2}} \le \dfrac{4\sqrt3}{9}$
Dấu bằng xảy ra khi $B=C,A = 2\arcsin\dfrac{1}{3}$
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