\[{\left( {\frac{{HA}}{{BC}} + \frac{{HB}}{{AC}} + \frac{{HC}}{{AB}}} \right)^2} \ge 3\sqrt {\frac{{HA}}{{BC}}.\frac{{HB}}{{AC}} + \frac{{HB}}{{AC}}.\frac{{HC}}{{AB}} + \frac{{HC}}{{AB}}.\frac{{HA}}{{BC}}} \](*)

Xét tam giác HAE và tam giác CAD có:

Chung \(\widehat A\)

\(\widehat {CDA} = \widehat {AEH}\)

⇒ ∆HAE ᔕ ∆CAD (g.g)

 ⇒ \(\frac{{HA}}{{CA}} = \frac{{AE}}{{AD}}\)

⇒\(\frac{{HA.HB}}{{CA.CB}} = \frac{{AE.HB}}{{AD.CB}} = \frac{{{S_{AHB}}}}{{{S_{ABC}}}}\)(1)

Tương tự ta có:

\(\frac{{HB.HC}}{{AB.AC}} = \frac{{{S_{AHC}}}}{{{S_{ABC}}}}\)(2)

\(\frac{{HC.HA}}{{BC.BA}} = \frac{{{S_{BHC}}}}{{{S_{ABC}}}}\)(3)

Cộng (1), (2), (3) theo từng vế ta có:

 \(\frac{{HA.HB}}{{CA.CB}} + \frac{{HB.HC}}{{AB.AC}} + \frac{{HC.HA}}{{BC.BA}} = \frac{{{S_{AHB}}}}{{{S_{ABC}}}} + \frac{{{S_{AHC}}}}{{{S_{ABC}}}} + \frac{{{S_{BHC}}}}{{{S_{ABC}}}} = 1\)(**)

Từ (*) và (**) ta có: \[{\left( {\frac{{HA}}{{BC}} + \frac{{HB}}{{AC}} + \frac{{HC}}{{AB}}} \right)^2} \ge 3\sqrt 1 = 3\]

Hay \(\frac{{HA}}{{BC}} + \frac{{HB}}{{AC}} + \frac{{HC}}{{AB}} \ge \sqrt 3 \).