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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
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 $x=y=z=1$
theo bđt cauchy schwarz ta có
\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)
Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$Khi đó BĐT đã cho trở thành:$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$Mặt khác ta có:$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$
CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$Từ $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$
Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:
$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$
$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$
$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$
Khi đó BĐT đã cho trở thành:
$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$
Mặt khác ta có:
$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$
CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$
Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$
Từ $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$
Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$
\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)
\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)
Ta có \(abc\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y+1}{4}\ge2\sqrt{\dfrac{x^2}{4}}=x\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{z+1}+\dfrac{z+1}{4}\ge y\\\dfrac{z^2}{x+1}+\dfrac{x+1}{4}\ge z\end{matrix}\right.\)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}+\dfrac{x+y+z}{4}+\dfrac{3}{4}\ge x+y+z\)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow\dfrac{3\left(x+y+z\right)}{4}\ge\dfrac{9}{4}\)
\(\Rightarrow\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\ge\dfrac{3}{2}=1,5\) (2)
Từ (1) và (2)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge1,5\) (đpcm )
Dấu " = " xảy ra khi \(x=y=z=1\)
Sửa: =>\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\left(1\right)\)
+,3 = x + y + z \(\ge\) \(3\sqrt[3]{xyz}\Rightarrow xyz\le1\)
+, P \(\ge\) \(3\sqrt[3]{\dfrac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\sqrt[3]{\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\dfrac{x+y+z+3}{3}}=\dfrac{3}{2}\)