Question

cho a, b, c là số dương thỏa mãn a+b+c=1

CMR:

a²/b+b²/c+c²/a>=3(a²+b²+c²)

Mình cần gấp ạ !!

 

Answer 1

Trước hết, với \(a+b+c=1\) ta có:

\(a^2+b^2+c^2=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\)

\(=\left(a^3+ab^2\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+a^2b+b^2c+c^2a\)

\(\ge2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a\)

Hay \(a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)

Từ đó:

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\)

\(\ge\dfrac{3\left(a^2b+b^2c+c^2a\right)\left(a^2+b^2+c^2\right)}{a^2b+b^2c+c^2a}=3\left(a^2+b^2+c^2\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)