Lời giải:
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Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}+\frac{c}{\sqrt{1-c}}=\frac{a}{\sqrt{a+b+c-a}}+\frac{b}{\sqrt{a+b+c-b}}+\frac{c}{\sqrt{a+b+c-c}}\)
\(=\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}=\frac{a^2}{a\sqrt{b+c}}+\frac{b^2}{b\sqrt{c+a}}+\frac{c^2}{c\sqrt{a+b}}\)
\(\geq \frac{(a+b+c)^2}{a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}}(*)\)
Và:
\((a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b})^2\leq (a+b+c)(ab+ac+bc+ba+ca+cb)=2(a+b+c)(ab+bc+ac)\)
Áp dụng BĐT AM-GM : \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}\) (quen thuộc)
Do đó:\((a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b})^2\leq \frac{2}{3}(a+b+c)^3\)
\(\Rightarrow a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq \sqrt{\frac{2}{3}(a+b+c)^3}(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{\sqrt{\frac{2}{3}(a+b+c)^3}}=\sqrt{\frac{3}{2}(a+b+c)}=\sqrt{\frac{3}{2}}\)
Vậy \(P_{\min}=\sqrt{\frac{3}{2}}\Leftrightarrow a=b=c=\frac{1}{3}\)
