Gọi G' là trọng tâm tam giác ABC. Dễ c/m: S,G,G' thẳng hàng và $\frac{SG}{SG'}=\frac{3}{4}$
Ta có: $\frac{SG}{SG'}.\frac{SM}{SA}.\frac{SN}{SB}=\frac{V_{S.MNG}}{V_{S.ABG'}}=3.\frac{V_{S.MNG}}{V_{S.ABC}}$
tuơng tự suy ra:
$\frac{SG}{SG'}(\frac{SM}{SA}.\frac{SN}{SB}+\frac{SN}{SB}.\frac{SP}{SC}+\frac{SP}{SC}.\frac{SM}{SA})=3.\frac{V_{S.MNG}+V_{S.NPG}+V_{S.PAG}}{V_{S.ABC}}=3.\frac{V_{S.MNP}}{V_{S.ABC}}=3.\frac{SM.SN.SP}{SA.SB.SC}\Leftrightarrow SM.SN+SN.SP+SP.SA=4.SM.SN.SP\Rightarrow ĐPCM$
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