Có $cosA.cosB.cosC = cosA. \frac{1}2[cos(B-C)+cos(B+C)]$$= \frac{1}2. cosA.[cos(B-C)-cosA] (*)$
Vì $cos(B-C) \leq 1$
$\Rightarrow (*) \leq \frac{1}2cosA(1-cosA)$
Ap dụng BĐT cô-si
Có $\sqrt{cosA(1-cosA)} \leq \frac{cosA+1-cosA}2 = \frac{1}2$
$\Rightarrow cosA(1-cosA) \leq (\frac{1}2)^2=\frac{1}4$
$\Rightarrow (*) \leq \frac{1}2.\frac{1}4=\frac{1}8$
Xét $cos^3A+cos^3B+cos^3C \geq 3\sqrt[3]{cos^3A.cos^3B.cos^3C} \geq 3.\frac{1}8=\frac{3}8 (đpcm)$