Ta có:$(1-cosA)(1-cosB)(1-cosC)=8sin^2\frac{A}{2}.sin^2\frac{B}{2}.sin^2\frac{C}{2}$
Áp dụng BĐT Cô-si,ta có:
$sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}\leq (\frac{sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}}{3})^3\leq [sin(\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3})]^3=sin^3\frac{\pi}{6}$
$sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}\leq \frac{1}{8}$
$\Rightarrow 8sin^2\frac{A}{2}.sin^2\frac{B}{2}.sin^2\frac{C}{2}\leq \frac{1}{8}$ (đpcm)