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Từ giả thiết : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Rightarrow xy+yz+zx=xyz\)
Ta có : \(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Vì hai vế luôn dương nên ta bình phương hai vế được :
\(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\ge\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\)
Xét \(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\)
\(=\left(x+y+z\right)+\left(xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)
Xét \(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\)
\(=xyz+\left(x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
Suy ra : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge\)
\(\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) (*)
Mà theo bất đẳng thức Bunhiacopxki , ta có :
\(\sqrt{\left(x+yz\right)}.\sqrt{y+zx}\ge\sqrt{xy}+\sqrt{yz.zx}=\sqrt{xy}+z\sqrt{xy}\) (1)
\(\sqrt{y+zx}.\sqrt{z+xy}\ge\sqrt{yz}+x\sqrt{yz}\)(2)
\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)(3)
Cộng (1) , (2) và (3) theo vế ta được (*) đúng
Vậy bđt ban đầu được chứng minh.
Ta có \(ax^3=by^3=cz^3\Leftrightarrow\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=ax^2+by^2+cz^2\Leftrightarrow\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}+\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}+\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)Vậy \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
\(0< x,y,z< 4\)\(\Rightarrow\)\(\hept{\begin{cases}x\left(x-4\right)< 0\\y\left(y-4\right)< 0\\z\left(z-4\right)< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2< 4x\\y^2< 4y\\z^2< 4z\end{cases}\Leftrightarrow}\hept{\begin{cases}x^3>\frac{x^4}{4}\\y^3>\frac{y^4}{4}\\z^3>\frac{z^4}{4}\end{cases}}}\)
\(\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}>\sqrt[4]{\frac{x^4}{4}}+\sqrt[4]{\frac{y^4}{4}}+\sqrt[4]{\frac{z^4}{4}}=\frac{x+y+z}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{1}{\left(\sqrt{5}+2\right)^2}\)
\(\Rightarrow N=\frac{1}{\left(\sqrt{5}-2\right)^2}-\frac{3}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{2}{9+4\sqrt{5}}\)
\(=\frac{1}{9-4\sqrt{5}}+\frac{2}{9+4\sqrt{5}}-3=\frac{9+4\sqrt{5}+18-8\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=24-4\sqrt{5}\)
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+x^2y^2-1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=2005^2-1\)
\(\Rightarrow S=\pm\sqrt{2005^2-1}\)
c/
Giả sử \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
\(\Leftrightarrow\sqrt[3]{3+\sqrt[3]{3}}-\sqrt[3]{3}< \sqrt[3]{3}-\sqrt[3]{3-\sqrt[3]{3}}\)
\(\Leftrightarrow\frac{\sqrt[3]{3}}{\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}}< \frac{\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\) (1)
Ta có: \(\left\{{}\begin{matrix}\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}\\\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}>\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\end{matrix}\right.\)
Nên (1) đúng
Vậy BĐT ban đầu đúng