Question

Cho a, b, c là các số thực dương thỏa mãn a + b + c = 1.Tìm giá trị nhỏ nhất của biểu thức:

\(A=\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\)

Answer 1

\(A=\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\left(\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}\right)+3\\ \overset{AM-GM}{\ge}\dfrac{9}{2a-a^2+2b-b^2+2c-c^2}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ \ge\dfrac{9}{2\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{3}}+3\\ =\dfrac{9}{2\cdot1-\dfrac{1}{3}}+3=\dfrac{42}{5}\)

Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}2a-a^2=2b-b^2=2c-c^2\\a=b=c\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)