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\(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Leftrightarrow\begin{matrix}a+b+c=1\\a^4+b^4+c^4\ge abc\end{matrix}\) \(x,y,z\ne0\Rightarrow a,b,c\ne0\)
\(a^2+b^2+x^2\ge ab+bc+ac\) (*){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 \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\) \(\Rightarrow\) \(\dfrac{1}{x.x}+\dfrac{1}{y.y}+\dfrac{1}{z.z}=1\)
\(\Rightarrow\dfrac{1}{x.x.x}+\dfrac{1}{y.y.y}+\dfrac{1}{z.z.z}=1\)\(\Rightarrow\dfrac{1}{x.x.x.x}+\dfrac{1}{y.y.y.y}+\dfrac{1}{z.z.z.z}=1\Leftrightarrow\dfrac{1}{x^4}+\dfrac{1^{ }}{y^4}+\dfrac{1}{z^4}=1\)
\(\Rightarrow\)\(\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{z^4}\)>= \(\dfrac{1}{x.y.z}\)
từ x+y+z=a và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=a\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{xyz}\)
<=>(xy+yz+xz)(x+y+z)=xyz
Từ đó bạn nhân phá ngoặc rồi biến phương trình trên về dạng:
(x+y)(y+z)(z+x)=0
=> x=-y =>z=a
hoặc y=-z =>x=a
hoặc z=-x =>y=a.
Mik nghĩ vậy nhé!
\(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) \(\dfrac{\Rightarrow1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=a+b+c=0\)
cơ bản \(\left(a+b+c\right)=0\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow x.y.z\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{1}{abc}.\left(a^3+b^3+c^3\right)=\dfrac{1}{abc}\left(3abc\right)=3=>dpcm\Leftrightarrow dccm\)
Lời giải:
Ta có:
\(A=\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}=\frac{1}{x(x+1)}+\frac{1}{y(y+1)}+\frac{1}{z(z+1)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{y}-\frac{1}{y+1}+\frac{1}{z}-\frac{1}{z+1}\)
\(=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)(1)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x}+\frac{1}{1}\geq \frac{4}{x+1}\) và tương tự với các phân thức còn lại rồi cộng lại:
\(\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\geq 4\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(\Leftrightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\leq \frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\right)(2)\)
Từ (1); (2) suy ra \(A\geq \frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)\)
Mà theo BĐT Cauchy- Schwarz ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}=\frac{9}{3}=3\)
Do đó: \(A\geq \frac{3}{4}(3-1)=\frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=1\)
Do \(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
=> \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=y+\dfrac{1}{z}\Leftrightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\Leftrightarrow x-y=\dfrac{y-z}{yz}\\y+\dfrac{1}{z}=z+\dfrac{1}{x}\Leftrightarrow y-z=\dfrac{1}{x}-\dfrac{1}{z}\Leftrightarrow y-z=\dfrac{z-x}{xz}\\z+\dfrac{1}{x}=x+\dfrac{1}{y}\Leftrightarrow z-x=\dfrac{1}{y}-\dfrac{1}{x}\Leftrightarrow z-x=\dfrac{x-y}{xy}\end{matrix}\right.\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\dfrac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)x^2y^2z^2=\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\) hoặc \(x^2y^2z^2-1=0\)
=> x=y=z hoặc xyz=1 hoặc xyz=-1
Lời giải:
Sử dụng điều kiện \(xyz=1\):
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow \left(x-\frac{1}{x}\right)+(y+z)-\left(\frac{1}{y}+\frac{1}{z}\right)=0\)
\(\Leftrightarrow \frac{x^2-1}{x}+(y+z)-\frac{(y+z)}{yz}=0\)
\(\Leftrightarrow yz(x^2-1)+(y+z)-x(y+z)=0\)
\(\Leftrightarrow (x-1)(xyz+yz)-(y+z)(x-1)=0\)
\(\Leftrightarrow (x-1)(1+yz)-(y+z)(x-1)=0\)
\(\Leftrightarrow (x-1)(yz+1-y-z)=0\)
\(\Leftrightarrow (x-1)(y-1)(z-1)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\)
Nghĩa là ít nhất một trong ba số có giá trị bằng 1 (đpcm)