a) MA.BC + MB.CA + MC .AB  ≥ 4SABC

Kẻ BE, CF vuông góc với AM.

Ta có:

MA.BC = MA.(BP + CP) ≥ MA.(BE + CF) = 2S_ABM + 2S_CAM

Tương tự:

MB.CA ≥ 2S_BCM + 2S_ABM

MC.AB ≥ 2S_CAM + 2S_BCM

Suy ra:

MA.BC + MB.CA + MC.AB ≥ 2(2S_BCM + 2S_ABM + 2 S_CAM) = 4S_ABC

Dấu “=” xảy ra khi M là trực tâm

b) S_PQR = S_ABC – S_AQR – S_BPQ – S_CRQ

Đặt \(\frac{{AQ}}{{QB}} = x;\frac{{BP}}{{PC}} = y;\frac{{CR}}{{RA}} = z\)

⇒ \(\frac{{AQ}}{{AB}} = \frac{x}{{x + 1}};\frac{{AR}}{{AC}} = \frac{1}{{z + 1}}\)

\[\frac{{{S_{AQR}}}}{{{S_{ACB}}}} = \frac{{AQ}}{{AB}}.\frac{{AR}}{{AC}} = \frac{x}{{\left( {x + 1} \right)\left( {z + 1} \right)}}\]

Tương tự: \[\frac{{{S_{BPQ}}}}{{{S_{ACB}}}} = \frac{y}{{\left( {y + 1} \right)\left( {x + 1} \right)}}\]

\[\frac{{{S_{CRP}}}}{{{S_{CBA}}}} = \frac{z}{{\left( {z + 1} \right)\left( {y + 1} \right)}}\]

⇒ \[\frac{{{S_{AQR}}}}{{{S_{ACB}}}} + \frac{{{S_{BPQ}}}}{{{S_{ACB}}}} + \frac{{{S_{CRP}}}}{{{S_{CBA}}}} = \frac{z}{{\left( {z + 1} \right)\left( {y + 1} \right)}} + \frac{y}{{\left( {y + 1} \right)\left( {x + 1} \right)}} + \frac{x}{{\left( {x + 1} \right)\left( {z + 1} \right)}}\]

\[\frac{{{S_{PQR}}}}{{{S_{ABC}}}} = \frac{{xyz + 1}}{{\left( {x + 1} \right)\left( {z + 1} \right)\left( {y + 1} \right)}}\]

Theo định lý Ceva có: xyz = 1 nên

\[\frac{{{S_{PQR}}}}{{{S_{ABC}}}} = \frac{{1 + 1}}{{\left( {x + 1} \right)\left( {z + 1} \right)\left( {y + 1} \right)}} = \frac{{1 + 1}}{{1 + xy + yz + zx + x + y + z + 1}}\]

Do xy + yz + zx + x + y + z\( \ge 6\sqrt[6]{{xyz}} = 6\)

Suy ra: \[\frac{{{S_{PQR}}}}{{{S_{ABC}}}} \le \frac{2}{8} = \frac{1}{4} \Rightarrow {S_{PQR}} \le \frac{1}{4}{S_{ABC}}\]

Dấu “=” xảy ra khi M là trọng tâm.