Áp dụng BĐT:$x^{3}+y^{3}\geq xy(x+y)$ (x,y>0)đpcm$\Leftrightarrow$$\Sigma \frac{abc}{a^{3}+b^{3}+abc} \leq \Sigma \frac{abc}{ab(a+b+c)} =\Sigma \frac{c}{a+b+c} =1$Dấu''='' xra$\Leftrightarrow$a=b=c
Áp dụng BĐT:$x^{3}+y^{3}\geq xy(x+y)$ (x,y>0)đpcm$\Leftrightarrow$$\Sigma \frac{abc}{a^{3}+b^{3}+abc} \leq \Sigma \frac{abc}{ab(a+b+c)} =\Sigma \frac{c}{a+b+c} =1$
Áp dụng BĐT:$x^{3}+y^{3}\geq xy(x+y)$ (x,y>0)đpcm$\Leftrightarrow$$\Sigma \frac{abc}{a^{3}+b^{3}+abc} \leq \Sigma \frac{abc}{ab(a+b+c)} =\Sigma \frac{c}{a+b+c} =1$
Dấu''='' xra$\Leftrightarrow$a=b=c