\(H=\frac{a^2}{2017a^2+a}+\frac{b^2}{2017b^2+b}+\frac{c^2}{2017c^2+c}\ge\frac{\left(a+b+c\right)^2}{2017\left(a^2+b^2+c^2\right)+\left(a+b+c\right)}\)
\(H\ge\frac{\left(a+b+c\right)^2}{2017.\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)}=\frac{1}{\frac{2017}{3}+1}=\frac{3}{2020}\)
\(\Rightarrow H_{max}=\frac{3}{2020}\) khi \(a=b=c=\frac{1}{3}\)
Bài dưới bị ngược dấu.
\(H=\frac{a}{2017a+1}+\frac{b}{2017b+1}+\frac{c}{2017c+1}\)
\(\Rightarrow 2017H=\frac{2017a}{2017a+1}+\frac{2017b}{2017b+1}+\frac{2017c}{2017c+1}=1-\frac{1}{2017a+1}+1-\frac{1}{2017b+1}+1-\frac{1}{2017c+1}\)
\(=3-\left(\frac{1}{2017a+1}+\frac{1}{2017b+1}+\frac{1}{2017c+1}\right)\)
Áp dụng BĐT Cauchy Schwarz:
\(\frac{1}{2017a+1}+\frac{1}{2017b+1}+\frac{1}{2017c+1}\geq \frac{9}{2017(a+b+c)+3}=\frac{9}{2020}\)
\(\Rightarrow 2017H=3-\left(\frac{1}{2017a+1}+\frac{1}{2017b+1}+\frac{1}{2017c+1}\right)\leq 3-\frac{9}{2020}=\frac{6051}{2020}\)
\(\Rightarrow H\leq \frac{3}{2020}\)
Vậy \(H_{\max}=\frac{3}{2020}\Leftrightarrow a=b=c=\frac{1}{3}\)
