b4) VT+7=\frac{b+c}{a}+2+\frac{2a+c}{b}+1+\frac{4(a+b)}{a+c}+4 =\frac{2a+b+c}{a}+\frac{2a+b+c}{b}+\frac{4(2a+b+c)}{a+c} =(a+b+a+c)(\frac{1}{a}+\frac{1}{b}+\frac{4}{a+c} \geq (1+1+2)^{2}=16 \Rightarrow VT\geq 9dấu "="\Leftrightarrow a=b=c
b4) VT+7=
\frac{b+c}{a}+2+\frac{2a+c}{b}+1+\frac{4(a+b)}{a+c}+4 =
\frac{2a+b+c}{a}+\frac{2a+b+c}{b}+\frac{4(2a+b+c)}{a+c} =$(a+b+a+c)(\frac{1}{a}+\frac{1}{b}+\frac{4}{a+c}
) \geq (1+1+2)^{2}=16
\Rightarrow VT\geq 9
dấu "="\Leftrightarrow a=b=c$