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Thế $a=2RsinA,b=2RsinB,c=2RsinC$ và $2p=a+b+c$ ta được$\frac{2R\left ( sinAcosA+sinBcosB+sinCcosC \right )}{2R\left ( sin^2A+sin^2B+sin^2C \right )}=\frac{a+b+c}{9R}$$\Leftrightarrow \frac{sin2A+sin2B+sin2C}{sin^2A+sin^2B+sin^2C}=\frac{4\left ( sinA+sinB+sinC \right )}{9}$biến đổi $sin2A+sin2B+sin2C=2sin(A+B)cos(A-B)+2sinCcosC=2sinC[cos(A-B)-cos(A+B)]=4sinAsinBsinC$$9sinAsinBsinC=(sin^2A+sin^2B+sin^2C)(sinA+sinB+sinC\geq 9\sqrt[3]{sin^3A+sin^3B+sin^3C} $$\Leftrightarrow sinA=sinB=sinC$$ \Leftrightarrow a=b=c$

Thế $a=2RsinA,b=2RsinB,c=2RsinC$ và $2p=a+b+c$ ta được$\frac{2R\left ( sinAcosA+sinBcosB+sinCcosC \right )}{2R\left ( sin^2A+sin^2B+sin^2C \right )}=\frac{a+b+c}{9R}$$\Leftrightarrow \frac{sin2A+sin2B+sin2C}{sin^2A+sin^2B+sin^2C}=\frac{4\left ( sinA+sinB+sinC \right )}{9}$biến đổi $sin2A+sin2B+sin2C=2sin(A+B)cos(A-B)+2sinCcosC=2sinC[cos(A-B)-cos(A+B)]=4sinAsinBsinC$$9sinAsinBsinC=(sin^2A+sin^2B+sin^2C)(sinA+sinB+sinC\geq 9\sqrt[3]{sin^3Asin^3B.sin^3C} $$\Leftrightarrow sinA=sinB=sinC$$ \Leftrightarrow a=b=c$