Bất đẳng thức cần chứng minh tương đương:

\(\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+zx}+\sqrt{z\left(x+y+z\right)+xy}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\). (1)

Theo bđt Bunhiakowski:

\(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\).

Tương tự: \(\sqrt{\left(y+z\right)\left(y+x\right)}\ge y+\sqrt{zx}\); \(\sqrt{\left(z+x\right)\left(z+y\right)}\ge z+\sqrt{xy}\).

Cộng vế với vế và kết hợp với gt x + y + z = 1 ta có (1) đúng.

Vậy ta có đpcm.

\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\)

Tương tự:

\(\sqrt{y+zx}\ge y+\sqrt{zx}\) ; \(\sqrt{z+xy}\ge z+\sqrt{xy}\)

Cộng vế với vế:

\(VT\ge\left(x+y+z\right)+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=...\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

\(P=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+xz}+\sqrt{z\left(x+y+z\right)+xy}\)

\(P=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)

\(P\le\dfrac{1}{2}\left(x+y+x+z\right)+\dfrac{1}{2}\left(x+y+y+z\right)+\dfrac{1}{2}\left(x+z+y+z\right)\)

\(P\le2\left(x+y+z\right)=2\)

\(P_{max}=2\) khi \(x=y=z=\dfrac{1}{3}\)

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

\(\Rightarrow xyz\le1\)

\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)

Ta co:

\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)

\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)

\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)

\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow A\ge xy+yz+zx\)

Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))

Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)

\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)

\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)

\(\ge xy+yz+zx\)

Đẳng thức xảy ra khi x = y = z = 1

\(\sqrt[3]{yz\cdot1}\le\frac{y+z+1}{3};\sqrt[3]{xz\cdot1}\le\frac{x+z+1}{3};\sqrt[3]{yx\cdot1}\le\frac{y+x+1}{3}\)

Nên \(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{y+x+1}\right)\)\(=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=B\)

\(B\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x+y+z}\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3\ge xy+yz+zx\)

do \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3=x^2+y^2+z^2;xy+yz+zx\le x^2+y^2+z^2=3\)

Đặt \(A=\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\)

Ta có:

\(x^2+xy+yz+zx=x+xyz=x\left(x+yz\right)\)

\(\Rightarrow\frac{x\left(x+yz\right)}{x}=\frac{x^2+xy+yz+zx}{x}\)

\(\Leftrightarrow x+yz=\frac{x^2+xy+yz+zx}{x}=\frac{\left(x^2+xy\right)+\left(yz+zx\right)}{x}=\frac{\left(x+z\right)\left(x+y\right)}{x}\)

\(\Rightarrow\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\)

Vì x, y, z >0 nên áp dụng bất đẳng thức Bunhiacopxki cho 2 số dương, ta được:

\(\left(x+y\right)\left(x+z\right)\ge\left(\sqrt{x^2}.+\sqrt{yz}\right)^2\)

\(\Rightarrow\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\)

\(\Rightarrow\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{\sqrt{x}}\)

Do đó \(\sqrt{x+yz}\ge\frac{x+\sqrt{yz}}{\sqrt{x}}\left(1\right)\)

Chứng minh tương tự, ta được:

\(\sqrt{y+xz}\ge\frac{y+\sqrt{xz}}{\sqrt{y}}\left(2\right)\)

Chứng minh tương tự, ta được:

\(\sqrt{z+xy}\ge\frac{z+\sqrt{xy}}{\sqrt{z}}\left(3\right)\)

Từ (1), (2) và (3), ta được:

\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\)\(\ge\frac{x+\sqrt{yz}}{\sqrt{x}}+\frac{y+\sqrt{zx}}{\sqrt{y}}+\frac{z+\sqrt{xy}}{\sqrt{z}}\)

\(\Leftrightarrow A\ge\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)

\(\Leftrightarrow A\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{yz+zx+xy}{\sqrt{xyz}}\)

 \(\Leftrightarrow A\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xyz}{\sqrt{xyz}}\)(vì \(xy+yz+zx=xyz\))

\(\Leftrightarrow A\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\)(điều phải chứng minh).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\xy+yz+zx=xyz\end{cases}}\Leftrightarrow x=y=z=3\)

Vậy với x, y, z là các số thực dương thỏa mãn xy + yz + zx =xyz thì:

\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\).

\(\)

\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

\(\sqrt{3x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)

Thay 2020=x+y+z vao mẫu đc

\(\frac{xy}{\sqrt{xy+zx+zy+z^2}}=\frac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\frac{xy}{2}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)(Cauchy)

Làm tương tự mấy cái kia sau đó ghép mấy cái cũng mẫu lại là ra

\(\Sigma\left(\frac{xy}{\sqrt{xy+2020z}}\right)=\Sigma\left[\frac{xy}{\sqrt{xy+z\left(x+y+z\right)}}\right]=\Sigma\left[\frac{xy}{\sqrt{\left(y+z\right)\left(z+x\right)}}\right]\)

\(=\Sigma\left[\sqrt{\frac{xy}{y+z}\cdot\frac{xy}{z+x}}\right]\le\Sigma\left[\frac{1}{2}\cdot\left(\frac{xy}{y+z}+\frac{xy}{z+x}\right)\right]\)

\(=\frac{1}{2}\cdot\left(\frac{xy}{y+z}+\frac{xy}{z+x}+\frac{yz}{x+y}+\frac{yz}{z+x}+\frac{zx}{x+y}+\frac{zx}{y+z}\right)\)

\(=\frac{1}{2}\cdot\left[\frac{x\left(y+z\right)}{y+z}+\frac{y\left(z+x\right)}{z+x}+\frac{z\left(x+y\right)}{x+y}\right]\)

\(=\frac{1}{2}\cdot\left(x+y+z\right)=\frac{2020}{2}=1010\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{2020}{3}\)