Giải: Kẻ đường cao AD,BE,CF. Đặt $S_{1}=S_{AFE},S_{2}=S_{BFD},S_{3}=S_{CED}$
Ta có:
$\Delta AEB \sim \Delta AFC(g.g)$⇒$\dfrac{AE}{AF}=\dfrac{AB}{AC}$
⇒$\Delta AEF \sim \Delta ABC(g.g)$⇒$\sqrt{\dfrac{S_1}{S}}=\dfrac{AE}{AB}=cosA$
Mặt khác,$ \sqrt{\dfrac{S_1}{S}}=\sqrt{\dfrac{AE.AF.sinA}{AB.AC.sinA}}=\sqrt{\dfrac{AE.AF}{AB.AC}}≤\dfrac{1}{2}(\dfrac{AF}{AB}+\dfrac{AE}{AC})$
⇒$cosA≤\dfrac{1}{2}(\dfrac{AE}{AB}+\dfrac{AF}{AC})(1)$(BĐT cô si)
Tương tự, $cosB≤\dfrac{1}{2}(\dfrac{FB}{AB}+\dfrac{BD}{BC})(2)$,$cosC≤\dfrac{1}{2}(\dfrac{DC}{BC}+\dfrac{CE}{AC})(3)$
Cộng (1)(2)(3) theo vế, ta được:
$cosA+cosB+cosC≤\dfrac{1}{2}(\dfrac{AE}{AB}+ \dfrac{AF}{AC}+\dfrac{FB}{AB}+ \dfrac{BD}{BC}+ \dfrac{DC}{BC}+ \dfrac{CE}{AC})=\dfrac{3}{2}$(đpcm)
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