\(Q=\frac{a}{b+2020-a}+\frac{b}{c+2020-b}+\frac{c}{a+2020-c}\)
\(Q=\frac{a}{b+a+b+c-a}+\frac{b}{c+a+b+c-b}+\frac{c}{a+a+b+c-c}\)
\(Q=\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\)
Áp dụng BĐT Cauchy-Schwarz:
\(Q=\frac{a^2}{a\cdot\left(2b+c\right)}+\frac{b^2}{b\cdot\left(2c+a\right)}+\frac{c^2}{c\cdot\left(2a+b\right)}\ge\frac{\left(a+b+c\right)^2}{3\cdot\left(ab+bc+ca\right)}\ge\frac{3\cdot\left(ab+bc+ca\right)}{3\cdot\left(ab+bc+ca\right)}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{2020}{3}\)
2020a hay là 2020-a vậy???
Nếu đề như vậy thì thay 2020 vô các mẫu đc
\(\frac{a}{2b+a}=\frac{a^2}{2ab+a^2}\)
Tương tự sau đó cosi swat là ra nha
