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Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\)
Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)
\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)
Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Dấu " = " xảy ra <=> x=y=z=1
Áp dụng bất đẳng thức Cacuhy - Schwarz
\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)
Áp dụng bất đẳng thức Cacuchy Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)
\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)
\(VT=\Sigma_{cyc}\frac{2\sqrt{x}}{x^3+y^2}\le\Sigma_{cyc}\frac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\Sigma_{cyc}\frac{1}{\sqrt{x^2y^2}}=\Sigma_{cyc}\frac{1}{xy}\)
\(=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\) (áp dụng BĐT quen thuộc \(ab+bc+ca\le a^2+b^2+c^2\))
Đẳng thức xảy ra khi x = y = z = 1
Sửa đề : \(\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{x^2z^2}}=\frac{2}{xz}\end{cases}}\)
\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)
Từ (1) và (2) :
\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)
Chúc bạn học tốt !!!
Đầu tiên CM BDT :
\(1+x^3+y^3\ge xy"x+y+z"\)
\(\Leftrightarrow x^3+y^3\ge xy"x+y"\)" do \(xyz=1\)"
\(\Leftrightarrow"x+y""x^2+y^2-xy"-xy"x+y"\ge0\)
\(\Leftrightarrow"x+y""x-y"^2\ge0\)
BDT luôn đúng theo gt
\(\Rightarrow\sqrt{"1+x^3+y^3"}\ge\sqrt{xy"x+y+z"}\)
\(\Rightarrow\sqrt{\frac{"1+x^3+y^3}{xy}}\ge\sqrt{\frac{"x+y+z"}{xz}}\)
Tương tự
\(\Rightarrow\sqrt{\frac{"1+z^3+y^3}{zy}}\ge\sqrt{\frac{"x+y+z"}{zy}}\)
\(\sqrt{\frac{"1+x^3+y^3"}{xz}}\ge\sqrt{\frac{"x+y+z"}{xz}}\)
\(\Rightarrow VT\ge\sqrt{"x+y+z"}.\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\)
AD BDT Cauchy cho các số > 0
\(x+y+z\ge3\). \(\sqrt[3]{xyz}=3\)
\(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\ge\frac{3}{\sqrt[3]{xyz}}=3\)
\(\Rightarrow VT\ge\sqrt{3}.3=3\sqrt{3}=VP\)
\(\Rightarrow VT\ge VP\)
\(\Rightarrow DPCM\)
Vậy Dấu \(= khi x=y=z=1\)
P/s: Thay dấu noặc kép thành ngọc đơn nha, Ko chắc đâu
Côsi: \(\sqrt{x\left(y+z\right)}=\frac{1}{2\sqrt{2}}.2.\sqrt{2x}.\sqrt{y+z}\le\frac{1}{2\sqrt{2}}\left(2x+y+z\right)\)
\(\Rightarrow\frac{1}{\sqrt{x\left(y+z\right)}}\ge\frac{2\sqrt{2}}{2x+y+z}\)
Tương tự các cái kia.
\(\Rightarrow VT\ge2\sqrt{2}\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)
\(\ge2\sqrt{2}.\frac{9}{2x+y+z+2y+z+x+2z+x+y}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
Từ giả thiết:\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c\)\(\Rightarrow ab+bc+ca=1\)
Ta có:\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)\(=\sqrt{\frac{1}{1+x^2}}+\sqrt{\frac{1}{1+y^2}}+\sqrt{\frac{1}{1+z^2}}\)
\(=\sqrt{\frac{\frac{1}{x}}{\frac{1}{x}+x}}+\sqrt{\frac{\frac{1}{y}}{\frac{1}{y}+y}}+\sqrt{\frac{\frac{1}{z}}{\frac{1}{z}+z}}\)\(=\sqrt{\frac{a}{a+\frac{1}{a}}}+\sqrt{\frac{b}{b+\frac{1}{b}}}+\sqrt{\frac{c}{c+\frac{1}{c}}}\)
\(=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)
Đến đây:\(\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{a^2+ab+bc+ca}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{a+c}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
Tương tự:\(\frac{b}{\sqrt{b^2+1}}\le\frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right);\frac{c}{\sqrt{c^2+1}}\le\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
Cộng 3 bất đẳng thức lại ta có điều phải chứng minh :))
\(\sqrt{z}=\sqrt{x}+\sqrt{y}\Rightarrow z=x+y+2\sqrt{xy}\Rightarrow x+y-z=-2\sqrt{xy}\)
\(\sqrt{y}=\sqrt{z}-\sqrt{x}\Rightarrow y=x+z-2\sqrt{zx}\Rightarrow z+x-y=2\sqrt{zx}\)
\(\sqrt{x}=\sqrt{z}-\sqrt{y}\Rightarrow x=y+z-2\sqrt{yz}\Rightarrow y+z-x=2\sqrt{yz}\)
\(\frac{1}{y+z-x}+\frac{1}{z+x-y}+\frac{1}{x+y-z}=\frac{1}{2}\left(\frac{1}{\sqrt{zx}}+\frac{1}{\sqrt{yz}}-\frac{1}{\sqrt{xy}}\right)\)
\(=\frac{1}{2}.\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\sqrt{xyz}}=0\)