Lời giải:
Ta có:
\(\text{VT}=a-\frac{ab(a+b)}{a^2+ab+b^2}+b-\frac{bc(b+c)}{b^2+bc+c^2}+c-\frac{ca(c+a)}{c^2+ca+a^2}\)
\(=a+b+c-\left(\frac{ab(a+b)}{a^2+ab+b^2}+\frac{bc(b+c)}{b^2+bc+c^2}+\frac{ca(c+a)}{c^2+ca+a^2}\right)\)
Áp dụng BĐT AM-GM:
\(\text{VT}\geq a+b+c-\left(\frac{ab(a+b)}{2ab+ab}+\frac{bc(b+c)}{2bc+bc}+\frac{ca(c+a)}{2ac+ac}\right)\)
\(\Leftrightarrow \text{VT}\geq a+b+c-\frac{2}{3}(a+b+c)=\frac{a+b+c}{3}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
a) \(bdt\Leftrightarrow\dfrac{2}{3}\left(a-b\right)^2\ge0\) (đúng). \("="\Leftrightarrow a=b\)
Câu b
một cách khác
\(A=\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\\ =\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+a^2c}\\ \ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+a^2c+ac^2}\\ =\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^3+a^2b+a^2c\right)+\left(b^3+ab^2+b^2c\right)+\left(c^3+c^2a+c^2b\right)}\\ =\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}\\ =\dfrac{a^2+b^2+c^2}{a+b+c}\)
Điều cần CM :
\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{a+b+c}\ge\dfrac{a+b+c}{3}\\ \Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\\ \Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\left(luon;dung\right)\)
=> đpcm
