Khờ Đz xin giải bài cuốiai cũng có :√a2−ab+b2=√(a+b)2−3ab≥√(a+b)2−3(a+b)24=a+b2(∗)Áp Dụng BĐT $\frac{1}{x+y}\leq \frac{1}{4}\left ( \frac{1}{a} +\frac{1}{b}\right )$$(*) \Rightarrow A\leq 2\left ( \frac{1}{a+b} + \frac{1}{b+c}+\frac{1}{c+a}\right )\leq \frac{1}{2}\left ( \frac{2}{a}+\frac{2}{b}+\frac{2}{c} \right )=3 \Leftrightarrow a=b=c=1$
Khờ Đz xin giải bài cuốiai cũng có :
√a2−ab+b2=√(a+b)2−3ab≥√(a+b)2−3(a+b)24=a+b2(∗)Áp Dụng BĐT $\frac{1}{x+y}\leq \frac{1}{4}\left ( \frac{1}{
x} +\frac{1}{
y}\right )$$(*) \Rightarrow A\leq 2\left ( \frac{1}{a+b} + \frac{1}{b+c}+\frac{1}{c+a}\right )\leq \frac{1}{2}\left ( \frac{2}{a}+\frac{2}{b}+\frac{2}{c} \right )=3 \Leftrightarrow a=b=c=1$