_ Chứng minh VT <2 .
Với a,b,c > 0, ta có:
\(a< a+b\Rightarrow\dfrac{a}{a+b}< 1=\dfrac{c}{c}\Rightarrow\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\) (1)
\(b< b+c\Rightarrow\dfrac{b}{b+c}< 1=\dfrac{a}{a}\Rightarrow\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\) (2)
\(c< c+a\Rightarrow\dfrac{c}{c+a}< 1=\dfrac{b}{b}\Rightarrow\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\) (3)
Từ (1) , (2) và (3), Cộng vế theo vế ta có:
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\)(*)
_Chứng minh VP > 2.
Theo BĐT Cô-si, ta có:
\(\sqrt{\dfrac{b+c}{a}.1}\le\left(\dfrac{b+c}{a}+1\right):2=\dfrac{b+c+a}{2a}\)
Do vậy : \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Tương tự:\(\sqrt{\dfrac{b}{a+c}}\ge\dfrac{2b}{a+b+c},\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế theo vế
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu ''='' xảy ra \(\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\)
\(\Rightarrow a+b+c=0\) (trái với g/t a,b,c >0)
Vậy đẳng thức khong xảy ra dấu ''=''
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\) (**)
Từ (*) và (**) \(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)
