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

\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{ac+bc+c^2+ab}}=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\)

\(tt\Rightarrow2\text{ lần biểu thức}=2\sqrt{\frac{bc}{\left(b+a\right)\left(c+a\right)}}+2\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+2\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)

\(\le\frac{b}{b+a}+\frac{c}{c+a}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+b}\left(\sqrt{ab}\le\frac{a+b}{2}\right)=3\Rightarrow dpcm\)