Question

 Cho ba số thực dương a, b, c . Chứng minh rằng: 

\(\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a++b}\ge a+b+c\)

Answer 1

\(\dfrac{a^2+bc}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)-a\left(b+c\right)}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}-a\)

\(\Rightarrow VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}-\left(a+b+c\right)\)

Mặt khác áp dụng \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

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

\(\Rightarrow VT\ge2\left(a+b+c\right)-\left(a+b+c\right)=a+b+c\) (đpcm)