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Em thử ạ.
a) ĐK: \(x\ge3;y\ge5;z\ge4\)
\(PT\Leftrightarrow\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-4}+\frac{4}{\sqrt{x-3}}+\frac{9}{\sqrt{y-5}}+\frac{25}{\sqrt{z-4}}=20\)
Ta có (theo BĐT AM-GM): \(\sqrt{x-3}+\frac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\frac{4}{\sqrt{x-3}}}=2.2=4\)
Tương tự:\(\sqrt{y-5}+\frac{9}{\sqrt{y-5}}\ge2.3=6\)
\(\sqrt{z-4}+\frac{25}{\sqrt{z-4}}\ge2.5=10\)
Cộng theo vế 3 BĐT trên được \(VT\ge20\)
Xảy ra đẳng thức khi \(\sqrt{x-3}=\frac{4}{\sqrt{x-3}}\Leftrightarrow x-3=4\Leftrightarrow x=7\)
Tương tự mấy cái kia ta cũng có \(y=14;z=29\)
Vậy..
Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)
ta có \(x^4+y^4\ge2x^2y^2\); \(y^4+z^4\ge2y^2z^2\);\(z^4+x^4\ge2z^2x^2\)
==> \(2\left(x^4+y^4+z^4\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)\)
<=> \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\)
mặt khác \(x^2y^2+y^2z^2\ge2xy^2z\)
\(y^2z^2+z^2x^2\ge2xyz^2\)
\(z^2x^2+x^2y^2\ge2x^2yz\)
==> \(2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge2xyz\left(x+y+z\right)=2xyz\)( vì x+y+z=1)
==> \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\)
dấu ''='' xảy ra khi x=y=z mà x+y+z=1 ==> x=y=z=1/3
vậy \(\left(x;y;z\right)=\left(\frac{1}{3};\frac{1}{3};\frac{1}{3}\right)\)
Bài 1 : ĐK : \(x>3\) ; \(y>5\) ; \(z>4\)
\(\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-4}=20-\dfrac{4}{\sqrt{x-3}}-\dfrac{9}{\sqrt{y-5}}-\dfrac{25}{\sqrt{z-4}}\)
\(\Leftrightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)=20\)
Theo BĐT Cô - Si cho hai số không âm ta có :
\(\left\{{}\begin{matrix}\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\dfrac{4\sqrt{x-3}}{\sqrt{x-3}}}=2\sqrt{4}=4\\\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge2\sqrt{\dfrac{9\sqrt{y-5}}{\sqrt{y-5}}}=2\sqrt{9}=6\\\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge2\sqrt{\dfrac{25\sqrt{z-4}}{\sqrt{z-4}}}=2\sqrt{25}=10\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)\ge20\)
\(\Rightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)=20\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=4\\y-5=9\\z-4=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=14\\z=29\end{matrix}\right.\left(TM\right)\)
Vậy \(x=7\) ; \(y=14\) ; \(z=29\)