Chứng minh bdt tổng quát:
[TEX]\frac{cos A}{x}+\frac{cos B}{y}+\frac{cos C}{z}\leq \frac{x^2+y^2+z^2}{2xyz}\\\\
\Rightarrow x^2+y^2+z^2\geq 2yzcosA+2xzcosB+2xycosC\\\\
\Rightarrow x^2-2x(zcosB+ycosC)+y^2+z^2-2yzcosA\\\\
\Delta' = (zcosB+ycosC)^2-(y^2+z^2-2yzcosA)\\\\
=-z^2(sinB)^2-y^2(sinC)^2+2yzcosBcosC+2yzcosA\\\\
=-(zsinB-ysinC)^2+2yzcosBcosC-2yzsinBsinC+2yzcosA\\\\
=-(zsinB-ysinC)^2 +2yzcos(B+C)+2yzcosA\\\\
= -(zsinB-ysinC)^2\leq 0 \forall x,y,z
\Rightarrow [/TEX] BDT được chứng minh
Aps dụng cho x=3,y=4,z=5 ta được [TEX]\frac{cos A}{3}+\frac{cos B}{4}+\frac{cos C}{5}\leq \frac{3^2+4^2+5^2}{2*3*4*5}\\\\=\frac{5}{12}[/TEX] (DPCM)