Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{a(b+c)}+\frac{\left(\frac{1}{b}\right)^2}{b(a+c)}+\frac{\left(\frac{1}{c}\right)^2}{c(a+b)}\)
\(\geq \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{a(b+c)+b(a+c)+c(a+b)}=\frac{(ab+bc+ac)^2}{2(ab+bc+ac)}=\frac{ab+bc+ac}{2}\) (thay $1=abc$)
Mà theo BĐT AM-GM:
\(ab+bc+ac\geq 3\sqrt[3]{(abc)^2}=3\). Do đó:
\(P\geq \frac{ab+bc+ac}{2}\geq \frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Cách khác:
Áp dụng BĐT AM-GM:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=\frac{1}{a}=\frac{abc}{a}=bc\)
Tương tự:
\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq ab\)
Cộng theo vế các BĐT trên ta có:
\(P+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)
\(\Rightarrow P\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{(abc)^2}}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
