Đặt \(x=a;y=\frac{b}{2};z=\frac{c}{3}\left(x,y,z>0\right)\) và\(x+y+z=xyz\)
Khi đó ta có: \(B=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{\sqrt{x^2+1}}=\sqrt{\frac{xyz}{x^2\left(x+y+z\right)+xyz}}\le\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}\le\frac{y}{2\left(x+y\right)}+\frac{z}{2\left(x+z\right)}\)
Tương tự có: \(\frac{1}{\sqrt{1+y^2}}\le\frac{x}{2\left(x+y\right)}+\frac{z}{2\left(y+z\right)};\frac{1}{\sqrt{1+z^2}}\le\frac{x}{2\left(x+z\right)}+\frac{y}{2\left(y+z\right)}\)
\(\Rightarrow B\le\frac{x+y}{2\left(x+y\right)}+\frac{x+z}{2\left(x+z\right)}+\frac{y+z}{2\left(y+z\right)}=\frac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\Rightarrow\hept{\begin{cases}a=\sqrt{3}\\b=2\sqrt{3}\\c=3\sqrt{3}\end{cases}}\)
