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Ta giả sử 3 số đều =2
=>\(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)(Đúng)
=>đpcm
P/s : nhanh gọn lẹ :))
Đặt \(A=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\)
Không mất tính tổng quát giả sử:
\(\frac{1}{x+1}< \frac{1}{y+1}< \frac{1}{z+1}\)
Ta có
+) \(A>\frac{3}{1+x}\Leftrightarrow1>\frac{3}{1+x}\)
\(\Leftrightarrow\frac{1}{3}>\frac{1}{x+1}\Leftrightarrow x+1>3\)
<=> x>2(1)
+) \(A< \frac{3}{1+z}\Leftrightarrow1< \frac{3}{1+z}\Leftrightarrow\frac{1}{3}< \frac{1}{1+z}\Leftrightarrow1+z< 3\Leftrightarrow x< 2\)(2)
Từ (1) (2) => ĐPCM
ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2006}\) (x;y;z khác 0)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)(vì x+y+z=2006)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-\left(x+y+z\right)}{\left(x+y+z\right).z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{\left(x+y+z\right).z}\)
\(\Leftrightarrow-\left(x+y\right)xy=\left(x+y\right)\left(xz+yz+z^2\right)\) (vì x;y;z khác 0)
\(\Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y=0 hoặc y+z=0 hoặc z+x=0
mà x+y+z=2006 nên
z=2006 hoặc x=2006 hoặc y=2006
=> đpcm
\(x^2-xy+y^2=\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\left(x+y\right)\)
Tương tự ta cũng có: \(\sqrt{x^2-xz+z^2}=\frac{1}{2}\left(x+z\right)\)
Suy ra \(\sqrt{x^2-xy+y^2}+\sqrt{x^2-xz+z^2}\ge\frac{1}{2}\left(2x+y+z\right)=1\)
Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{2}\).
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow x+y=0\) hoặc \(y+z=0\) hoặc \(z+x=0\)
=> ...............................................
Gọi cái vế trái là P.
Áp dụng BĐT Cauchy Ta có
\(P.\frac{2}{\sqrt{3}}=\sqrt{\left(x+1\right).\frac{4}{3}}+\sqrt{\left(y+1\right).\frac{4}{3}}+\sqrt{\left(z+1\right).\frac{4}{3}}\)
\(\le\frac{x+1+\frac{4}{3}}{2}+\frac{y+1+\frac{4}{3}}{2}+\frac{z+1+\frac{4}{3}}{2}=\frac{x+y+z+7}{2}=\frac{8}{2}=4\)
Do đó \(P\le4:\frac{2}{\sqrt{3}}=2\sqrt{3}< \frac{7}{2}\)
Ta có \(x+y+z=2\Leftrightarrow\frac{1}{x+y+z}=\frac{1}{2}\)
Mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\)
Suy ra \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\Leftrightarrow\left(x+y\right)\left[\frac{xy+xz+yz+z^2}{xy\left(xz+yz+z^2\right)}\right]=0\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)}{xy\left(xz+yz+z^2\right)}=0\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: x=-y\(\Leftrightarrow x+y+z=2\Leftrightarrow\left(-y\right)+y+z=2\Leftrightarrow z=2\)
TH2: y=-z\(\Leftrightarrow x+y+z=2\Leftrightarrow x+\left(-z\right)+z=2\Leftrightarrow x=2\)
TH3: z=-x\(\Leftrightarrow x+y+z=2\Leftrightarrow x+y+\left(-x\right)=2\Leftrightarrow y=2\)
Suy ra có ít nhất một trong ba số x,y,z bằng 2
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+y+z}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{y+z}{x\left(x+y+z\right)}+\frac{y+z}{yz}=0\)
\(\Leftrightarrow\left(y+z\right)\left(\frac{1}{x\left(x+y+z\right)}+\frac{1}{yz}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+z=0\\\frac{1}{x\left(x+y+z\right)}+\frac{1}{yz}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x^2+xy+xz=-yz\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2+xy+xz+yz=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+y\right)\left(x+z\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(2-z\right)\left(2-y\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right.\)