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\(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)
<=> x+y = 0 hoặc x+z=0 hoặc z+y=0
<=> x = -y hoặc x = -z hoặc z = -y
\(\Rightarrow P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)
Ta có:
\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A được:
\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)(do xy+yz+xz=1)
=>Đpcm
Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.
mk nhớ lm bài tương tự thế này r` bn chịu khó mở ra xem lại ở đây olm.vn/?g=page.display.showtrack&id=424601&limit=260, ấn vào chữ Trang tiếp theo để tìm thêm nhé
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
\(\left\{{}\begin{matrix}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)
\(\Leftrightarrow x+y=0\) hoặc \(x+z=0\) hoặc \(z+y=0\)
\(\Leftrightarrow x=-y\) hoặc \(x=-z\) hoặc z=-y
\(\Rightarrow P\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)
Chúc bạn học tốt !!
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Lời giải:
Bạn phải thêm đk \(x,y,z\) là những số không âm.
Đặt \(\frac{x}{2008}=\frac{y}{2009}=\frac{z}{2010}=k(k\geq 0)\Rightarrow x=2008k; y=2009k; z=2010k\)
Khi đó:
\(z-x=2010k-2008k=2k\)
\(\left\{\begin{matrix} x-y=2008k-2009k=-k\\ y-z=2009k-2010k=-k\end{matrix}\right.\)
\(\Rightarrow 2\sqrt{(x-y)(y-z)}=2\sqrt{(-k)(-k)}=2\sqrt{k^2}=2|k|=2k\)
Do đó: \(z-x=2\sqrt{(x-y)(y-z)}\)
Ta có đpcm.