\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)

\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)

\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

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

\(\Rightarrow\frac{a}{b-c}=-\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab-c^2+ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Tương tự:

\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)};\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Cộng lại:

\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

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bđt<=>\(S_a\left(a-b\right)^2+S_b\left(b-c\right)^2+S_c\left(c-a\right)^2\ge0\)

with \(S_a=\frac{1}{2\left(a^2+b^2\right)}-\frac{c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(S_b=\frac{1}{2\left(b^2+c^2\right)}-\frac{a}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(S_c=\frac{1}{2\left(c^2+a^2\right)}-\frac{b}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

cần cm \(S_a+S_c;S_b+S_c>0\)

lại có:\(S_a+S_c=\frac{1}{2}\left(\frac{1}{a^2+b^2}+\frac{1}{c^2+a^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}\)

\(>\frac{1}{2}\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(c+a\right)^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}>0\)

cmtt=>q.e.d