Từ giả thiết \(x+y+z=xyz=\frac{1}{xy}\)\(=\frac{1}{yz}\)\(=\frac{1}{zx}\)\(=1\)
Đặt \(\frac{1}{x}\)\(=a,\frac{1}{y}\)\(=b,\frac{1}{z}\)\(=c=ab+bc+ca=1\)
Ta có :
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)\(=\sqrt{\frac{1}{\sqrt{1+x^2}}+}+\sqrt{\frac{1}{\sqrt{1+y^2}}+\sqrt{\frac{1}{\sqrt{1+z^2}}}}\)
\(=\sqrt{\frac{1}{x}+x}+\sqrt{\frac{1}{y}+y}+\sqrt{\frac{1}{z}+z}=\sqrt{\frac{a}{a+\frac{1}{a}}}+\sqrt{\frac{b}{b+\frac{1}{b}}}\)\(+\sqrt{\frac{c}{c+\frac{1}{c}}}\)
\(=\frac{a}{\sqrt{a^2}+1}\)\(+\frac{b}{\sqrt{b^2}+1}\)\(+\frac{c}{\sqrt{c^2}+1}\)
Đến đây :
\(\frac{a}{\sqrt{a^2}+1}\)\(=\frac{a}{\left(a^2+ab+bc+ca\right)}\)\(=\frac{a}{\sqrt{\left(a+b\right)}\left(a+c\right)}\)
\(=\sqrt{\frac{a}{a+b}}\)\(\cdot\frac{a}{a+c}\)\(< \frac{1}{2}\)\(\left(\frac{b}{b+a}+\frac{b}{b+c}\right);\frac{c}{\sqrt{c^2}+1}\)\(< \frac{1}{2}\)\(\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
ộng 3 bất đẳng thức lại ta có điều phải chứng minh
