Lời giải:
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} a^2+b^2\geq 2ab\\ b^2+c^2\geq 2bc\\ c^2+a^2\geq 2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\geq ab+bc+ac\)
\(\Leftrightarrow (a+b+c)^2\geq3(ab+bc+ac)\Leftrightarrow ab+bc+ac\leq \frac{1}{3}\)
(do \(a+b+c=1\) )
Do đó, \(\frac{1}{ab+bc+ac}\geq \frac{1}{\frac{1}{3}}=3\)
Vậy \(P_{\min}=3\Leftrightarrow a=b=c=\frac{1}{3}\)
Giải
Áp dụng BĐT AM-MG :
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) \(\Leftrightarrow\left(a+b+c\right)^2\ge3.\left(ab+bc+ac\right)\Leftrightarrow ab+bc+ac\le\dfrac{1}{3}\)
Do đó \(\dfrac{1}{ab+bc+ac}\ge\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy Pmin = 3 \(\Leftrightarrow a=b=c\dfrac{1}{3}\)
