Khờ Đz xin giải bài cuốiai cũng có :$\sqrt{a^2-ab+b^2}=\sqrt{(a+b)^2-3ab}\geq \sqrt{(a+b)^2-\frac{3(a+b)^2}{4}}=\frac{a+b}{2} (*)$Áp Dụng BĐT $\frac{1}{x+y}\leq \frac{1}{4}\left ( \frac{1}{x} +\frac{1}{y}\right )$$(*) \Rightarrow A\leq 2\left ( \frac{1}{a+b} + \frac{1}{b+c}+\frac{1}{c+a}\right )\leq \frac{1}{2}\left ( \frac{2}{a}+\frac{2}{b}+\frac{2}{c} \right )=3 \Leftrightarrow a=b=c=1$
Khờ Đz xin giải bài cuốiai cũng có :$\sqrt{a^2-ab+b^2}=\sqrt{(a+b)^2-3ab}\geq \sqrt{(a+b)^2-\frac{3(a+b)^2}{4}}=\frac{a+b}{2} (*)$Áp Dụng BĐT $\frac{1}{x+y}\leq \frac{1}{4}\left ( \frac{1}{a} +\frac{1}{b}\right )$$(*) \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 :$\sqrt{a^2-ab+b^2}=\sqrt{(a+b)^2-3ab}\geq \sqrt{(a+b)^2-\frac{3(a+b)^2}{4}}=\frac{a+b}{2} (*)$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$