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![](https://rs.olm.vn/images/avt/0.png?1311)
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 $x=y=z=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)
\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)
\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow abc=1\)
\(P=\dfrac{a^2bc}{b+c}+\dfrac{ab^2c}{c+a}+\dfrac{abc^2}{a+b}=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)
\(VT\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)
Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)
\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)
Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)
\(\Rightarrow A\ge\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+,3 = x + y + z \(\ge\) \(3\sqrt[3]{xyz}\Rightarrow xyz\le1\)
+, P \(\ge\) \(3\sqrt[3]{\dfrac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\sqrt[3]{\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\dfrac{x+y+z+3}{3}}=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy, ta có:
\(VT\ge3\sqrt[3]{\dfrac{x^2.y^2.z^2}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Ta có: xyz=1 và x,y,z >0
\(\Rightarrow x\le1\Rightarrow x+1\le2\Rightarrow\dfrac{1}{x+1}\ge\dfrac{1}{2}\)
Tương tự \(\dfrac{1}{y+1}\ge\dfrac{1}{2}\)
\(\dfrac{1}{z+1}\ge\dfrac{1}{2}\)
\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi x=y=z=1
Do \(x;y>0\) ta biến đổi tương đương:
\(\dfrac{x+y}{2}\ge\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}\Leftrightarrow\dfrac{x+y}{2}\ge\dfrac{2}{\dfrac{x+y}{xy}}\)
\(\Leftrightarrow\dfrac{x+y}{2}\ge\dfrac{2xy}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh, dấu "=" xảy ra khi \(x=y\)