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$A=\frac{a^2-a+a+b}{b+c} + \frac{b^2-b+b+c}{c+a} + \frac{c^2-c+c+a}{a+b}$Mà $a+b+c=1$ nên ta có:$A=\frac{a^2-a+a+b}{1-a} + \frac{b^2-b+b+c}{1-b} + \frac{c^2-c+c+a}{1-c}$$A=-a-b-c + \frac{a+b}{1-b} + \frac{b+c}{1-b} + \frac{c+a}{1-c}$Lại có $a+b+c=1$ :$A=-1 + \frac{a+b}{b+c} + \frac{b+c}{c+a} + \frac{c+a}{a+b}$$A=-1+\frac{(a+b)^2}{b^2+ab+bc+ca}+\frac{(b+c)^2}{c^2+ab+bc+ca}+\frac{(c+a)^2}{a^2+ab+bc+ca}$Áp dụng bđt Bunhia Copxki ta có:$A\geq -1 + \frac{(2a+2b+2c)^2}{a^2+b^2+c^2+3ab+3bc+3ca}$$A \geq -1 + \frac{4.(a+b+c)^2}{(a+b+c)^2+ab+bc+ca}$Mà $3(ab+bc+ca)\leq(a+b+c)^2$ ( Biến đổi tương đương)$\Rightarrow ab+bc+ca\leq \frac{(a+b+c)^2}{3}$$\Rightarrow A\geq -1 + \frac{4.(a+b+c)^2}{(a+b+c)^2+\frac{(a+b+c)^2}{3}}$$A \geq -1 + 3 = 2$Dấu bằng xảy ra $\Leftrightarrow$ $a=b=c=1$

$A=\frac{a^2-a+a+b}{b+c} + \frac{b^2-b+b+c}{c+a} + \frac{c^2-c+c+a}{a+b}$Mà $a+b+c=1$ nên ta có:$A=\frac{a^2-a+a+b}{1-a} + \frac{b^2-b+b+c}{1-b} + \frac{c^2-c+c+a}{1-c}$$A=-a-b-c + \frac{a+b}{1-b} + \frac{b+c}{1-b} + \frac{c+a}{1-c}$Lại có $a+b+c=1$ :$A=-1 + \frac{a+b}{b+c} + \frac{b+c}{c+a} + \frac{c+a}{a+b}$$A=-1+\frac{(a+b)^2}{b^2+ab+bc+ca}+\frac{(b+c)^2}{c^2+ab+bc+ca}+\frac{(c+a)^2}{a^2+ab+bc+ca}$Áp dụng bđt Bunhia Copxki ta có:$A\geq -1 + \frac{(2a+2b+2c)^2}{a^2+b^2+c^2+3ab+3bc+3ca}$$A \geq -1 + \frac{4.(a+b+c)^2}{(a+b+c)^2+ab+bc+ca}$Mà $3(ab+bc+ca)\leq(a+b+c)^2$ ( Biến đổi tương đương)$\Rightarrow ab+bc+ca\leq \frac{(a+b+c)^2}{3}$$\Rightarrow A\geq -1 + \frac{4.(a+b+c)^2}{(a+b+c)^2+\frac{(a+b+c)^2}{3}}$$A \geq -1 + 3 = 2$Dấu bằng xảy ra $\Leftrightarrow$ $a=b=c=\frac{1}{3}$