Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)

\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)

Tương tự cũng có:

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

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

Cộng vế với vế ta được:

 \(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)

Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)

hay

\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)

\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)

Cộng vế:

\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)

\(A=\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\)

\(\dfrac{\sqrt{x^3+y^3+1}}{xy}=\dfrac{\sqrt{x^3+y^3+xyz}}{xy}\ge\dfrac{\sqrt{xy\left(x+y\right)+xyz}}{xy}=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}\ge\dfrac{\sqrt{xy.3^3\sqrt{xyz}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)

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

\(\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge\dfrac{\sqrt{3}}{\sqrt{zx}}\)

\(\Rightarrow A\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.xz}}}=3\sqrt{3}.\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

Bài này cũng dễ mà:

Áp dụng BĐT Cô-si, ta có:

\(y+z+1\ge3\sqrt[3]{yz}\)

\(\Rightarrow\)\(\dfrac{y+z+1}{3}\ge\sqrt[3]{yz}\)

\(\Rightarrow\)\(\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{3x}{y+z+1}\)

\(\Rightarrow\)\(\sum\dfrac{x}{\sqrt[3]{yz}}\ge\sum\dfrac{3x}{y+z+1}\)

Mà \(\sum\dfrac{3x}{y+z+1}=\sum\dfrac{3x^2}{xy+xz+x}\)

Áp dụng BĐT Cauchy -Schwaz:

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

Mà:

\(xy+yz+xz\le x^2+y^2+z^2\)(BĐT phụ)

\(\Rightarrow\)\(2\left(xy+yz+xz\right)\le2\left(x^2+y^2+z^2\right)=6\)

Áp dụng BĐT Bunhicopski:

\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)

\(\Rightarrow x+y+z\le3\)

\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le6+3=9\)

\(\Rightarrow\)\(\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{3\left(x+y+z\right)^2}{9}\ge\dfrac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\left(ĐPCM\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\)x=y=z=1

@Lightning Farron vào thể hiện đẳng cấp đi anh zai :))

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Ta có:

\(VT=\dfrac{1}{x^2+yz}+\dfrac{1}{y^2+xz}+\dfrac{1}{z^2+xy}\le\dfrac{1}{2x\sqrt{yz}}+\dfrac{1}{2y\sqrt{xz}}+\dfrac{1}{2z\sqrt{xy}}\)

\(\Rightarrow VT\le\dfrac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\le\dfrac{\dfrac{x+y}{2}+\dfrac{y+z}{2}+\dfrac{x+z}{2}}{2xyz}=\dfrac{x+y+z}{2xyz}\)

Dấu "=" xảy ra khi \(x=y=z\)

Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)

  \(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)

Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)

                         \(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

Cộng vế với vế ta có:

\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)

\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)