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Ta có: \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge9xyz\)
\(VT=\dfrac{x}{1+yz}+\dfrac{y}{1+xz}+\dfrac{z}{1+xy}\)
\(=\dfrac{x^2}{x+xyz}+\dfrac{y^2}{y+xyz}+\dfrac{z^2}{z+xyz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3xyz}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\dfrac{\left(x+y+z\right)\left(xy+yz+xz\right)}{3}}\)
\(=\dfrac{3\left(x+y+z\right)}{4}\). Cần chứng minh:
\(\dfrac{3\left(x+y+z\right)}{4}\ge\dfrac{3\sqrt{3}}{4}\Leftrightarrow x+y+z\ge\sqrt{3}\)
BĐT cuối đúng vì \(x+y+z\ge\sqrt{3\left(xy+yz+xz\right)}=\sqrt{3}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Ps: nospoiler
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+z^2\ge xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2zx+z^2\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\left(luôn.đúng\right)\)
Dấu "=' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng
Dấu "=" xảy ra khi \(x=y=z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\frac{\left(yz\right)^2}{x^2yz\left(y+z\right)}+\frac{\left(zx\right)^2}{xy^2z\left(z+x\right)}+\frac{\left(xy\right)^2}{xyz^2\left(x+y\right)}\)
\(VT=\frac{2\left(yz\right)^2}{xy+xz}+\frac{2\left(zx\right)^2}{xy+yz}+\frac{2\left(xy\right)^2}{xz+yz}\)
\(VT\ge\frac{2\left(xy+yz+zx\right)^2}{2\left(xy+yz+zx\right)}=xy+yz+zx\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt[3]{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\frac{xy}{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)
Tương tự: \(\sqrt{\frac{yz}{yz+x}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{x+z}\right)\) ; \(\sqrt{\frac{zx}{zx+y}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{z}{y+z}\right)\)
Cộng vế với vế ta có đpcm
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(yz\le\frac{1}{2}\left(y^2+z^2\right)=\frac{1}{2}\left(3-x^2\right)\)
\(\Rightarrow3-yz\ge3-\frac{1}{2}\left(3-x^2\right)=\frac{3}{2}+\frac{1}{2}x^2\)
\(\Rightarrow\frac{x}{3-yz}\le\frac{x}{\frac{3}{2}+\frac{1}{2}x^2}=\frac{2x}{x^2+3}\)
Làm tương tự và cộng lại ta có: \(VT\le2\left(\frac{x}{x^2+3}+\frac{y}{y^2+3}+\frac{z}{z^2+3}\right)\)
Ta sẽ chứng minh: với mọi \(0< x^2< 3\) ta luôn có: \(\frac{x}{x^2+3}\le\frac{x^2+3}{16}\)
Thật vậy, BĐT tương đương:
\(16x\le\left(x^2+3\right)^2\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+9\right)\ge0\) (luôn đúng)
Tương tự: \(\frac{y}{y^2+3}\le\frac{y^2+3}{16}\) ; \(\frac{z}{z^2+3}\le\frac{z^2+3}{16}\)
Cộng vế với vế:
\(VT\le2.\frac{x^2+y^2+z^2+9}{16}=\frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)