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Link : https://123doc.org/document/3369350-ung-dung-cua-dinh-ly-viet.htm
Trang 2 nhé :33
Ta co:\(\Sigma\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}=\Sigma\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)Ta lai co:
\(\Sigma x+\Sigma\frac{1}{x}=\Sigma\left(x+\frac{1}{4x}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+\frac{3}{4}.\frac{9}{x+y+z}\ge3+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{2}\)
Vay \(P_{min}=\frac{15}{2}\)khi \(x=y=z=\frac{1}{2}\)
1.
a.
\(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+3}\)
b.
Theo de bai ta co:
\(A\ge0\left(DK:\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\right)\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+3}\ge0\)
\(\Leftrightarrow3\sqrt{x}\ge0\)
\(\Leftrightarrow x\ge0\)
Vay de \(A\ge0\)thi \(\left\{{}\begin{matrix}x\in\left[0;+\infty\right]\\x\ne9\end{matrix}\right.\)