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 "="$\Leftrightarrow a=b=c$
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 "="$\Leftrightarrow a=b=c$
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 "="$\Leftrightarrow a=b=c$