Lời giải:
Theo BĐT Cauchy Schwarz:
\(ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}\)
\(\Rightarrow a+b+c\geq 3\)
Áp dụng BĐT AM-GM:
\(A=a-\frac{ca}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}\)
\(=(a+b+c)-\left(\frac{ac}{c+a^2}+\frac{ab}{a+b^2}+\frac{bc}{b+c^2}\right)\)
\(\geq (a+b+c)-\left(\frac{ac}{2a\sqrt{c}}+\frac{ab}{2b\sqrt{a}}+\frac{bc}{2c\sqrt{b}}\right)\)
\(A\geq (a+b+c)-\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\)
Cũng theo BĐT AM-GM:
\(\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{a+1}{2}+\frac{b+1}{2}+\frac{c+1}{2}=\frac{a+b+c+1}{4}\)
\(\Rightarrow A\geq a+b+c-\frac{a+b+c+3}{4}=\frac{3}{4}(a+b+c)-\frac{3}{4}\geq \frac{3}{4}.3-\frac{3}{4}=\frac{3}{2}\)
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
