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\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)
\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)
\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)
\(D\le\dfrac{1}{2}\left(1+\dfrac{x}{1+yz}\right)+\dfrac{1}{2}\left(1+\dfrac{y}{1+zx}\right)+\dfrac{z}{2+2xy}\)
\(=1+\dfrac{x}{2\left(1+yz\right)}+\dfrac{y}{2\left(1+zx\right)}+\dfrac{z}{2\left(1+xy\right)}\)
Do \(0\le x;y;z\le1\)
\(\Rightarrow\left(1-x\right)\left(1-y\right)\ge0\Leftrightarrow xy+1\ge x+y\)
\(\Leftrightarrow2\left(xy+1\right)\ge xy+1+x+y\ge x+y+z\)
\(\Rightarrow\dfrac{z}{2\left(1+xy\right)}\le\dfrac{z}{x+y+z}\)
Tương tự: \(\dfrac{x}{2\left(1+yz\right)}\le\dfrac{x}{x+y+z}\) ; \(\dfrac{y}{2\left(1+zx\right)}\le\dfrac{y}{x+y+z}\)
Cộng vế:
\(P\le1+\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=2\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow ab+bc+ca=1\)
Đặt vế trái là P, ta có:
\(P=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)
\(P=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(P=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)
Lời giải:
Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)
Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)
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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)
\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)
\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)
\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)
Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)
Suy ra:
\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)
\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$