\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
\(P^2=2+2\sqrt{\left(a+b\right)\left(b+c\right)}+2\sqrt{\left(b+c\right)\left(c+a\right)}\)
\(+2\sqrt{\left(a+b\right)\left(c+a\right)}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow2\sqrt{\left(a+b\right)\left(b+c\right)}\le a+2b+c\)
Tương tự ta có : \(\hept{\begin{cases}2\sqrt{\left(b+c\right)\left(c+a\right)}\le a+b+2c\\2\sqrt{\left(a+b\right)\left(a+c\right)}\le2a+b+c\end{cases}}\)
\(\Rightarrow P^2\le2+4\left(a+b+c\right)\)
\(\Rightarrow P\le\sqrt{6}\)
Vậy \(P_{Max}=\sqrt{6}\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
