Question

Cho a, b thỏa mãn \(a^2++b^2=2\)

Tìm M_min = \(\dfrac{a^3}{2016a+2017b}+\dfrac{b^3}{2017a+2016b}\)

Answer 1

\(M=\dfrac{a^3}{2016a+2017b}+\dfrac{b^3}{2017a+2016b}=\dfrac{a^4}{2016a^2+2017ab}+\dfrac{b^4}{2017ab+2016b^2}\)

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(M\ge\dfrac{\left(a^2+b^2\right)^2}{2016\left(a^2+b^2\right)+4034ab}=\dfrac{4}{4032+4034ab}\)

AM-GM: \(a^2+b^2\ge2ab\Leftrightarrow2ab\le2\Leftrightarrow ab\le1\Leftrightarrow4034ab\le4034\)

Hay: \(M\ge\dfrac{4}{4032+4034}=\dfrac{4}{8066}=\dfrac{2}{4033}\)