Question

\(\sqrt{\dfrac{a+b}{a+b+c}}+\sqrt{\dfrac{b+c}{a+b+c}}+\sqrt{\dfrac{c+a}{a+b+c}}\le\sqrt{6}\left(a,b,c>0\right)\)

Answer 1

\(VT\le\sqrt{3\left(\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}\right)}=\sqrt{6}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Answer 2

Đặt `A=` biểu thức trên

Áp dụng bunhia:

`=>A^2<=(1+1+1)((a+b)/(a+b+c)+(b+c)/(a+b+c)+(c+a)/(a+b+c))`

`=>A^2<=3.2=6`

`=>A<=sqrt6(do \ A>=0)`

Dấu "=" `<=>a=b=c`