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\(\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)
\(=\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+\dfrac{1}{x}}+\dfrac{1}{1+\dfrac{1}{xy}+\dfrac{1}{y}}\)
\(=\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+1}+\dfrac{xy}{xy+1+x}=\dfrac{1+x+xy}{1+x+xy}=1\)
Áp dụng công thức a3+b3+c3=3abc
Bài làm
Đặt \(\dfrac{1}{x}\)= a, \(\dfrac{1}{y}\)= b, \(\dfrac{1}{z}\)= c
vì a+b+c = 0 nên a3+b3+c3=3abc
S= \(\dfrac{yz}{x^2}\)+ \(\dfrac{xz}{y^2}\)+ \(\dfrac{xy}{z^{ }2}\)
=\(\dfrac{xyz}{x^{ }3}\)+\(\dfrac{xyz}{y^{ }3}\)+\(\dfrac{xyz}{z^{ }3}\) = xyz(\(\dfrac{1}{x^3}\)+\(\dfrac{1}{y^{ }3}\)+\(\dfrac{1}{z^{ }3}\))
= xyz ( a3+b3+c3 )
= xyz \(\times\)3abc = xyz \(\times\) \(\dfrac{3}{xyz}\) = 3
\(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Leftrightarrow\begin{matrix}a+b+c=1\\a^4+b^4+c^4\ge abc\end{matrix}\) \(x,y,z\ne0\Rightarrow a,b,c\ne0\)
\(a^2+b^2+x^2\ge ab+bc+ac\) (*){cơ bản} \(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\ge\left(ab.ac\right)+\left(ab.bc\right)+\left(ac.bc\right)=abc\left(a+b+c\right)=abc\)
(*) bình phương hai vế
\(\Leftrightarrow a^4+b^4+c^4+2\left(ab\right)^2+2\left(ac\right)^2+2\left(bc\right)^2\ge\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge-\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]+2abc\ge-abc+2abc=abc=>dpcm\)Đẳng thức:
a=b=c=1/3=> x=y=z=3
ta co \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\) \(\Rightarrow\) \(\dfrac{1}{x.x}+\dfrac{1}{y.y}+\dfrac{1}{z.z}=1\)
\(\Rightarrow\dfrac{1}{x.x.x}+\dfrac{1}{y.y.y}+\dfrac{1}{z.z.z}=1\)\(\Rightarrow\dfrac{1}{x.x.x.x}+\dfrac{1}{y.y.y.y}+\dfrac{1}{z.z.z.z}=1\Leftrightarrow\dfrac{1}{x^4}+\dfrac{1^{ }}{y^4}+\dfrac{1}{z^4}=1\)
\(\Rightarrow\)\(\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{z^4}\)>= \(\dfrac{1}{x.y.z}\)
a: Thiếu vế phải rồi bạn
b: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow\left(x-y\right)^2>=0\)(luôn đúng)
\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=-\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}.\left(-\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
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