Từ giả thiết suy ra : \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Nên ta có : \(\frac{\sqrt{1+x^2}}{x}=\sqrt{\frac{1}{x^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}\le\frac{1}{2}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow y=z\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}\le\frac{1}{2}\left(\frac{4}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có :
\(\frac{1+\sqrt{1+y^2}}{y}\le\frac{1}{2}\left(\frac{1}{x}+\frac{4}{y}+\frac{1}{z}\right);\frac{1+\sqrt{1+z^2}}{z}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}\right)\)
Vậy ta có :
\(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Ta có :
\(\left(x+y+z\right)^2-3\left(xy+yz+xx\right)=...=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\ge0\)
Nên \(\left(x+y+x\right)^2\ge3\left(xy+yz+xx\right)\)
\(\Rightarrow\left(xyz\right)^2\ge3\left(xy+yz+xz\right)\Rightarrow3\frac{xy+yz+xz}{xyz}\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le xyz\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le xyz\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Chúc bạn học tốt !!
\(\frac{1+\frac{1}{2}.2.\sqrt{1+x^2}}{x}\le\frac{1+\frac{1}{4}\left(x^2+5\right)}{x}=\frac{x}{4}+\frac{9}{4x}\)
\(\Rightarrow VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4xyz}=\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4\left(x+y+z\right)}\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{3\left(x+y+z\right)^2}{4\left(x+y+z\right)}=x+y+z=xyz\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)