đặt \(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
\(=>A^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
\(=>A^2\le\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+a}\right)^2\right].3\)
\(=>A^2\le\left[2\left(a+b+c\right)\right]3=2.3=6\)
\(=>A\le\sqrt{6}\left(dpcm\right)\)
dấu"=" xảy ra<=>a=b=c=1/3
Ta có:\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2=\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)=3.2=6\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Dấu "=" xảy ra <=> a=b=c=1/3
