Cho x,y,z là các số thực thỏa mãn
(x-y)(x-z)=1 và y\(\ne\)z
Chứng minh rằng \(\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(y-z\right)^2}+\frac{1}{\left(z-x\right)^2}\)\(\ge4\)
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Đặt \(\left\{{}\begin{matrix}x-y=a\\x-z=b\end{matrix}\right.\) \(\Rightarrow z-y=a-b\) và \(ab=1\)
\(VT=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a-b\right)^2}=\frac{a^2+b^2}{a^2b^2}+\frac{1}{\left(a-b\right)^2}\)
\(VT=a^2+b^2+\frac{1}{\left(a-b\right)^2}=\left(a-b\right)^2+\frac{1}{\left(a-b\right)^2}+2ab=\left(a-b\right)^2+\frac{1}{\left(a-b\right)^2}+2\)
\(VT\ge2\sqrt{\frac{\left(a-b\right)^2}{\left(a-b\right)^2}}+2=4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-y\right)\left(x-z\right)=1\\\left(y-z\right)^2=1\end{matrix}\right.\)
\(\Sigma\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\Sigma\left(\dfrac{1}{9}.\dfrac{a^2\left(2+1\right)^2}{2a.\left(\Sigma a\right)+2a^2+bc}\right)\le\Sigma\left(\dfrac{1}{9}.\dfrac{4a^2}{2a\left(\Sigma a\right)}+\dfrac{1}{9}.\dfrac{a^2}{2a^2+bc}\right)\)
\(=\Sigma\left(\dfrac{1}{9}.\left(\dfrac{2a}{\Sigma a}+\dfrac{a^2}{2a^2+bc}\right)\right)=\dfrac{1}{9}\left(2+\Sigma\dfrac{a^2}{2a^2+bc}\right)\)
Cần chứng minh \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
<=> \(\Sigma\frac{bc}{2a^2+bc}\ge1\) (*)
Đặt (x;y;z) -------> \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Suy ra (*) <=> \(\Sigma\frac{x^2}{x^2+2xy}\ge1\Leftrightarrow\frac{\Sigma x^2}{\Sigma x^2}\ge1\) (đúng)
Vậy \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
Suy ra \(\Sigma\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\le\frac{1}{9}\left(2+\Sigma\frac{a^2}{2a^2+bc}\right)\le\frac{1}{9}\left(2+1\right)=\frac{1}{3}\)
Đẳng thức xảy ra <=> x = y = z = 1
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.
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) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Ta đặt \(\hept{\begin{cases}x+z=a\\y+z=b\end{cases}\Rightarrow ab=1}\)
\(BĐT\Leftrightarrow\frac{1}{\left(a-b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge4\)
Ta có
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{\left(a-\frac{1}{a}\right)^2}+a^2+\frac{1}{a^2}\)
\(=\frac{1}{\left(a-\frac{1}{a}\right)^2}+\left(a-\frac{1}{a}\right)^2+2\)
\(\ge2+2=4\)
Đặt \(\left\{{}\begin{matrix}x-y=a\\z-x=b\\y-z=c\end{matrix}\right.\) đề bài trở thành \(\left\{{}\begin{matrix}abc\ne0\\a+b+c=0\\ab=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\b=-\frac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{c^2}=\frac{1}{\left(a+b\right)^2}\\b^2=\frac{1}{a^2}\end{matrix}\right.\)
Ta cần chứng minh \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge4\)
\(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}=\frac{1}{a^2}+a^2+\frac{1}{\left(a-\frac{1}{a}\right)^2}\)
\(P=\left(a-\frac{1}{a}\right)^2+\frac{1}{\left(a-\frac{1}{a}\right)^2}+2\ge2\sqrt{\left(a-\frac{1}{a}\right)^2.\frac{1}{\left(a-\frac{1}{a}\right)^2}}+2=4\) (đpcm)