Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)

Ta có:

\(x^2+y^2\ge2xy\Rightarrow x^2+y^2-xy\ge xy\)

\(\Leftrightarrow\left(x+y\right)\left(x^2+y^2-xy\right)\ge xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3\ge xy\left(x+y\right)\)

\(\Rightarrow\frac{1}{x^3+y^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}=\frac{1}{x+y+z}.\frac{1}{xy}\)

Tương tự: \(\frac{1}{y^3+z^3+xyz}\le\frac{1}{x+y+z}.\frac{1}{yz}\) ;\(\frac{1}{z^3+x^3+xyz}\le\frac{1}{x+y+z}.\frac{1}{zx}\)

\(\Rightarrow\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{z^3+x^3+xyz}\)

\(\le\frac{1}{x+y+z}.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{x+y+z}{\left(x+y+z\right)xyz}=\frac{1}{xyz}\)

Dấu \(=\) xảy ra \(\Leftrightarrow x=y=z>0\)

AD BĐT X^3+Y^3>=XY(X+Y) LÀ RA

Có BĐT phụ:

\(a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Áp dụng

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\)

\(\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{zx\left(z+x\right)+xyz}\)

\(=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}\)

\(=\frac{1}{xyz}\)

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

\(x^4y+x^2y-x^2y=x^2y\left(x^2+1\right)-x^2y.\)

\(\hept{\begin{cases}\frac{x^2y\left(x^2+1\right)-x^2y}{\left(x^2+1\right)}=x^2y-\frac{x^2y}{\left(x^2+1\right)}\\\frac{y^2z\left(y^2+1\right)-y^2z}{\left(y^2+1\right)}=y^2z-\frac{y^2z}{\left(y^2+1\right)}\\\frac{z^2x\left(z^2+1\right)-z^2x}{\left(z^2+1\right)}=z^2x-\frac{z^2x}{\left(z^2+1\right)}\end{cases}}Vt\ge x^2y+y^2z+z^2x-\left(\frac{x^2y}{x^2+1}+\frac{y^2z}{y^2+1}+\frac{z^2x}{z^2+1}\right)\)

\(\hept{\begin{cases}x^2+1\ge2x\\y^2+1\ge2y\\z^2+1\ge2z\end{cases}\Leftrightarrow\hept{\begin{cases}-\frac{x^2y}{x^2+1}\ge\frac{x^2y}{2x}=\frac{xy}{2}\\\frac{y^2z}{2y}=\frac{yz}{2}\\\frac{z^2x}{2z}=\frac{xz}{2}\end{cases}\Leftrightarrow}VT\ge x^2y+y^2z+z^2x-\left(\frac{xy+yz+zx}{2}\right)}\)

\(x^2y+y^2z+z^2x\ge3\sqrt[3]{x^3y^3z^3}=3\)

\(VT\ge3-\frac{\left(xy+yz+zx\right)}{2}\)

t chỉ làm dc đến đây thôi :))

Từ \(VT\ge x^2y+y^2z+z^2x-\left(\frac{xy+yz+zx}{2}\right)\)ta có:

\(x^2y+x^2y+y^2z=x^2y+x^2y+\frac{y}{x}\ge3xy\)(áp dụng BĐT Cauchy)

Tương tự : \(y^2z+y^2z+z^2x\ge3yz\);   \(z^2x+z^2x+x^2y\ge3zx\)

Cộng vế theo vế suy ra : \(3\left(x^2y+y^2z+z^2x\right)\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2y+y^2z+z^2x\ge xy+yz+zx\)

\(\Leftrightarrow VT\ge\frac{xy+yz+zx}{2}\ge\frac{3\sqrt[3]{x^2y^2z^2}}{2}=\frac{3}{2}\)

Dấu '=' xảy ra khi x = y = z = 1

Do xyz=1. nên bđt cần chứng minh tường đương với

\(\frac{x^4}{x^3z+xz}+\frac{y^4}{y^3x+xy}+\frac{z^4}{z^3y+zy}\ge\frac{3}{2}\)

Theo BĐT Bunhiacopsky ta có:

\(\frac{x^4}{x^3z+xz}+\frac{y^4}{y^3x+xy}+\frac{z^4}{z^3y+zy}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^3z+xz+y^3x+xy+z^3y+zy}\)

Do vậy ta cần cm

\(\frac{\left(x^2+y^2+z^2\right)^2}{x^3z+xz+y^3x+xy+z^3y+zy}\ge\frac{3}{2}\)

\(\Leftrightarrow2\left(x^4+y^4+z^4\right)+4\left(x^2y^2+y^2z^2+z^2x^2\right)\ge3\left(x^3z+y^3x+z^3y\right)+3\left(xy+yz+xz\right)\)

BĐT trên là tổng của 3 BĐT sau:

\(1,x^2y^2+y^2z^2+z^2x^2\ge xy+yz+xz\)

\(2,x^4+y^4+z^4\ge x^3z+y^3x+z^3y\)

\(3,x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2\ge2\left(x^3z+y^3x+z^3y\right)\)

ta có bđt trên tương đương với

\(x^2\left(x-z\right)^2+y^2\left(y-x\right)^2+z^2\left(z-y\right)^2\ge0\)

Nhân 3 ở bđt đầu tiên rồi cộng vế theo vế các bđt ở dưới ta có đpcm

dấu "=" xảy ra khi x=y=z=1

Á nhầm nhaaa cái cuối cùng là cộng z2 đó

Ta có :

\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)

tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)

\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)

\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)

Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)

Ta có: \(\frac{1+\sqrt{1+x^2}}{x}=\frac{1+\sqrt{1\times\left(1+x^2\right)}}{x}\le\frac{1+\frac{1+1+x^2}{2}}{x}=\frac{2+\frac{x^2}{2}}{x}=\frac{2}{x}+\frac{x}{2}\)(Áp dụng bđt Cauchy ở chỗ \(\sqrt{1\times\left(1+x^2\right)}\)

Tương tự với b,c . Ta được VT\(\le\)\(\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)

\(\le\)\(\frac{x+y+z}{2}\)+ \(\frac{2xy+2yz+2xz}{xyz}\)= \(\frac{x+y+z}{2}\)+ \(\frac{4xy+4yz+4xz}{2xyz}\)= \(\frac{xyz}{2}+\frac{4xy+4yz+4xz}{2xyz}\)

Ta chứng minh được \(4xy+4yz+4xz\le\left(x+y+z\right)^2\)bằng phương pháp biến đổi tương đương

=> VT \(\le\)\(\frac{xyz}{2}+\frac{\left(x+y+z\right)^2}{2xyz}=\frac{xyz}{2}+\frac{\left(xyz\right)^2}{2xyz}=\frac{xyz}{2}+\frac{xyz}{2}=xyz\)(Điều phải cm)

Dấu = xảy ra <=>