\(\left(a,b,c\right)\rightarrow\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(\Rightarrow A=\sum\sqrt{\dfrac{1}{1+\left(\dfrac{x}{y}\right)^2}}=\sum\sqrt{\dfrac{y^2}{x^2+y^2}}=\sum\sqrt{\dfrac{y^2\left(x^2+z^2\right)}{\left(x^2+y^2\right)\left(x^2+z^2\right)}}\)
ÁP dụng Bunyakovsky:
\(\sum\sqrt{\dfrac{y^2\left(x^2+z^2\right)}{\left(x^2+y^2\right)\left(x^2+z^2\right)}}\le\sqrt{2\left(x^2y^2+y^2z^2+z^2x^2\right)\left(\sum\dfrac{1}{\left(x^2+y^2\right)\left(x^2+z^2\right)}\right)}\)
\(=\sqrt{2\left(x^2y^2+y^2z^2+z^2x^2\right).\dfrac{2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)}}\)
Cần chứng minh \(VT\le\dfrac{3}{\sqrt{2}}\)
\(\Leftrightarrow\left(x^2y^2+y^2z^2+z^2x^2\right)\left(x^2+y^2+z^2\right)\le\dfrac{9}{8}\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)\)
( đúng )
Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=1