Quay lại đáp án

a) Ta có $\dfrac{11a+9b}{a(a+b)}=\dfrac{2a+9(a+b)}{a(a+b)}=\dfrac{2}{a+b}+\dfrac{9}{a}$Suy ra $\dfrac{11a+9b}{a(a+b)}+\frac{11b+9c}{b(b+c)}+\dfrac{11c+9a}{c(c+a)} \ge 2\left (\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{a+b} \right )+9\left (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right )$$\ge 2.\frac{9}{2(a+b+c)}+9.\frac{9}{a+b+c}=9+81=90$Đẳng thức xảy ra $\iff a=b=c=1/3$.

a) Ta có $\dfrac{11a+9b}{a(a+b)}=\dfrac{2a+9(a+b)}{a(a+b)}=\dfrac{2}{a+b}+\dfrac{9}{a}$Suy ra $\dfrac{11a+9b}{a(a+b)}+\frac{11b+9c}{b(b+c)}+\dfrac{11c+9a}{c(c+a)} \ge 2\left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right )+9\left (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right )$$\ge 2.\frac{9}{2(a+b+c)}+9.\frac{9}{a+b+c}=9+81=90$Đẳng thức xảy ra $\iff a=b=c=1/3$.