Lời giải:
Bổ sung: Tìm min \(A=\frac{1}{a^2(b+c)}+\frac{1}{b^2(c+a)}+\frac{1}{c^2(a+b)}\)
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Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{(abc)^2}{a^2(b+c)}+\frac{(abc)^2}{b^2(c+a)}+\frac{(abc)^2}{c^2(a+b)}=\frac{(bc)^2}{b+c}+\frac{(ca)^2}{c+a}+\frac{(ab)^2}{a+b}\geq \frac{(bc+ca+ab)^2}{b+c+c+a+a+b}\)
hay \(A\geq \frac{(ab+bc+ac)^2}{2(a+b+c)}(*)\)
Áp dụng BĐT Cauchy:
\(a^2b^2+b^2c^2\geq 2ab^2c; b^2c^2+c^2a^2\geq 2abc^2; c^2a^2+a^2b^2\geq 2a^2bc\)
\(\Rightarrow 2(a^2b^2+b^2c^2+2c^2a^2)\geq 2a^2bc+2ab^2c+2abc^2\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c)\)
\(\Leftrightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)\Leftrightarrow (ab+bc+ac)^2\geq 3(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow A\geq \frac{3(a+b+c)}{2(a+b+c)}=\frac{3}{2}\)
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
