Question

cho a,b,c>0 ,\(a^2+b^2+c^2=1\).CMR

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\)

Dùng bunhia nhé mn.Giúp e với e cần gấp ạ !

 

 

Answer 1

Đặt vế trái BĐT là P

Ta có:

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

\(\Rightarrow P.\left(2ab+2bc+2ca\right)\ge1\)

\(\Rightarrow P\ge\dfrac{1}{2\left(ab+bc+ca\right)}\ge\dfrac{1}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\) (đpcm)

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

Answer 2