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cosA.cosB.cosC≤18 Thât vậy :
cosAcosBcosC−18=12[cos(A+B)+cos(A−B)]cosC−18
=12[−cos2C+cos(A−B)cosC−14]
=−12[(cosC−12cos(A−B))2+14−14cos2(A−B)]
=−12[(cosC−12cos(A−B))2+14sin2(A−B)]≤0
Suy ra cosA.cosB.cosC≤18
Dấu "=" xảy ra \Leftrightarrow \begin{cases}cosC=\frac{1}{2}cos(A-B) \\ sin(A-B)=0 \end{cases}
\Leftrightarrow A=B=C=\frac{\pi}{3} hay \Delta ABC đều
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