Question

cho a,b,c>0,a+b+c=3

 CMR P=\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge3\)

Answer 1

Ta có: \(a+b+c=3\)  

Áp dụng BĐT Cauchy - Schwarz ta có:

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

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

\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{3^2}{2\cdot3}=\dfrac{3}{2}\)

__________________

Nhắc lại BĐT Cauchy - Schwarz:

\(\dfrac{x^2_1}{a_1}+\dfrac{x^2_2}{a_2}+\dfrac{x^2_3}{a_3}+...+\dfrac{x^2_n}{a_n}\ge\dfrac{\left(x_1+x_2+...+x_n\right)^2}{a_1+a_2+...+a_n}\) 

(p/s: bạn xem lại để nhé !)