Question

cho\(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\).

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

Answer 1

Áp dụng BĐT Bunhiacốpxki dạng phân thức có

\(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge\dfrac{\left(a+b+c\right)^2}{a+2b^2+b+2c^2+c+2a^2}=\dfrac{9}{3+2\left(a^2+b^2+c^2\right)}\) (1)

Áp dụng BĐT Bunhiacốpxki có:

\(\left(a.1+b.1+c.1\right)^2\ge\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)

\(\Rightarrow9\ge3\left(a^2+b^2+c^2\right)\Rightarrow3\ge a^2+b^2+c^2\Rightarrow2\left(a^2+b^2+c^2\right)\le6\) (2)

Thay (2) vào (1) có \(\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+a^2}\ge\dfrac{9}{3+6}=1\) (đpcm)

Dấu = xảy ra khi a= b=c=1