Question

cho a,b,c duong va abc=1

cmr \(Q=\sqrt{\frac{a}{1+a+ab}}+\sqrt{\frac{b}{1+b+bc}}+\sqrt{\frac{c}{1+c+ca}}\le\sqrt{a+b+c}\)

Answer 1

Áp dụng bất đẳng thức bu nhi a ta có \(\left(x^2+y^2+z^2\right)3\ge\left(x+y+z\right)^2\)

Áp dụng ta có 

\(Q^2\le3\left(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}\right)\)

đặt \(M=\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}=\frac{a}{1+a+ab}+\frac{ab}{a+ab+abc}+\frac{abc}{ab+abc+â^2bc}\)

    \(=\frac{1}{a+ab+1}+\frac{a}{a+ab+1}+\frac{ab}{1+ab+1}=1\)

=> \(Q^2\le3\Rightarrow Q\le\sqrt{3}\)

mặt khác Áp dụng cô si ta có 

\(a+b+c\ge3\sqrt[3]{abc}=3\Rightarrow\sqrt{a+b+c}\ge\sqrt{3}\Rightarrow\sqrt{a+b+c}\ge Q\) (ĐPCM)

Answer 2

ta có:

\(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}=\frac{a}{abc+a+ab}+\frac{b}{1+b+bc}+\frac{bc}{b+bc+abc}\)

\(=\frac{1}{1+b+bc}+\frac{b}{1+b+bc}+\frac{bc}{1+b+bc}=1\)

ta có:

\(Q^2\le3\left(\frac{a}{1+a+ab}+\frac{b}{1+b+bc}+\frac{c}{1+c+ca}\right)=3\)

\(\Rightarrow Q\le\sqrt{3}=\sqrt{3\sqrt[3]{abc}}\le\sqrt{a+b+c}\left(Q.E.D\right)\)

dấu = xảy ra khi a=b=c=1