Bài 2:
Ta có `\sqrt{\frac{ab}{ab+c}}=\sqrt{\frac{ab}{ab+1.c}}=\sqrt{\frac{ab}{ab+ac+bc+c^2}}`
`=\sqrt{\frac{ab}{(a+b)(a+c)}`
Tương tự
`\sqrt{\frac{bc}{bc+a}}=\sqrt{\frac{bc}{(b+c)(a+c)}`
`\sqrt{\frac{ac}{ac+b}}=\sqrt{\frac{ac}{(a+b)(b+c)}`
`=>A=\sqrt{\frac{ab}{ab+c}}+\sqrt{\frac{bc}{bc+a}}+\sqrt{\frac{ac}{ac+b}`
`=\sqrt{\frac{ab}{(a+b)(a+c)}}+\sqrt{\frac{bc}{(b+c)(a+c)}}+\sqrt{\frac{ac}{(a+b)(b+c)}` `=>2A=2\sqrt{\frac{ab}{(a+b)(a+c)}}+2\sqrt{\frac{bc}{(b+c)(a+c)}}+2\sqrt{\frac{ac}{(a+b)(b+c)}`
Ta có `a,b,c>0`
`=>\frac{a}{a+b}+\frac{b}{a+c}>=2\sqrt{\frac{ab}{(a+b)(a+c)}`
`=>2\sqrt{\frac{ab}{(a+b)(a+c)}}<=\frac{a}{a+b}+\frac{b}{a+c}`
Tương tự
`2\sqrt{\frac{bc}{(b+c)(a+c)}}<=\frac{b}{b+c}+\frac{c}{a+c}`
`2\sqrt{\frac{ac}{(a+b)(b+c)}}<=\frac{a}{a+b}+\frac{b}{b+c}`
`=>2A<=\frac{a}{a+b}+\frac{b}{a+c}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{a}{a+b}+\frac{b}{b+c}`
`=>2A<=3`
`=>A<=\frac{3}{2}`(ĐPCM)
Dấu bằng xảy ra khi
`a=b=c=1/3`
