-Áp dụng BĐT Bunhiacopxski ta có:
\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{\left(1^2+1^2+1^2\right)\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+a}\right)^2\right]}=\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{3.2\left(a+b+c\right)}=\sqrt{3.2.3}=3\sqrt{2}\)
\(P=3\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}\sqrt{a+b}=\sqrt{b+c}=\sqrt{c+a}\\a+b+c=3\end{matrix}\right.\Leftrightarrow a=b=c=1\)
-Vậy \(P_{max}=3\sqrt{2}\)
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