Lời giải:

Ta có: \(x+y+z=xyz\Rightarrow x(x+y+z)=x^2yz\)

\(\Rightarrow x(x+y+z)+yz=x^2yz+yz\)

\(\Rightarrow (x+y)(x+z)=yz(x^2+1)\)

Do đó: \(\frac{1+\sqrt{x^2+1}}{x}=\frac{1+\sqrt{\frac{(x+y)(x+z)}{yz}}}{x}\leq \frac{1+\frac{1}{2}(\frac{x+y}{y}+\frac{x+z}{z})}{x}\) theo BĐT AM-GM:

Thực hiện tương tự với các phân thức khác ta suy ra:

\(\text{VT}\leq \frac{1+\frac{1}{2}(\frac{x+y}{y}+\frac{x+z}{z})}{x}+\frac{1+\frac{1}{2}(\frac{y+z}{z}+\frac{y+x}{x})}{y}+\frac{1+\frac{1}{2}(\frac{z+x}{x}+\frac{z+y}{y})}{z}\)

\(\text{VT}\leq 3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3(xy+yz+xz)}{xyz}\)

Mà theo AM-GM:

\(\frac{3(xy+yz+xz)}{xyz}\leq \frac{(x+y+z)^2}{xyz}=\frac{(xyz)^2}{xyz}=xyz\)

Do đó: \(\text{VT}\leq xyz\)

Ta có đpcm.

may báo cáo Tiép đi

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 \(x=y=z=\sqrt{3}\)

tick đi rồi làm cho

1) VT= \(\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xyz}{xyz+z+zx}\)

\(=\frac{1}{1+x+xy}+\frac{xy}{1+x+xy}+\frac{xyz}{z\left(x+xy+1\right)}\)

\(=\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy}\)

\(=\frac{1+x+xy}{1+x+xy}=1\)

Bài 2 giả thiết trên tử làm mell gì có bình phương, nếu có thì tính làm gì nữa :D, kết quả là 2016(x+y+z)

đề b2 sai