Question

\(Cho\)\(a;b;c>0\)\(TM\)\(a+b+c=1\)

Tìm GTLN của \(Q=\frac{\left(a+1\right)^2}{2a^2+\left(b+c\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(c+a\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(a+b\right)^2}\)

Answer 1

ta có \(Q=\frac{a^2+2a+1}{2a^2+\left(1-a\right)^2}+...\)

              \(=\frac{a^2+2a+1}{3a^2-2a+1}+...=\frac{1}{3}+\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}+...\)

              \(=1+\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}+\frac{\frac{8}{3}b+\frac{2}{3}}{3b^2-2b+1}+\frac{\frac{8}{3}c+\frac{2}{3}}{3c^2-2c+1}\)

mà \(3a^2-2a+1=3\left(a-\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

=>\(\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}\le\frac{\frac{8}{3}a+\frac{2}{3}}{\frac{2}{3}}=\frac{3}{2}\left(\frac{8}{3}a+\frac{2}{3}\right)=4a+1\)

tương tự mấy cái kia rồi + vào, ta có 

\(Q\le1+4\left(a+b+c\right)+3=8\)

dấu = xảy ra <=>a=b=c=1/3

^_^