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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 \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\) \(\Rightarrow\) \(\dfrac{1}{x.x}+\dfrac{1}{y.y}+\dfrac{1}{z.z}=1\)
\(\Rightarrow\dfrac{1}{x.x.x}+\dfrac{1}{y.y.y}+\dfrac{1}{z.z.z}=1\)\(\Rightarrow\dfrac{1}{x.x.x.x}+\dfrac{1}{y.y.y.y}+\dfrac{1}{z.z.z.z}=1\Leftrightarrow\dfrac{1}{x^4}+\dfrac{1^{ }}{y^4}+\dfrac{1}{z^4}=1\)
\(\Rightarrow\)\(\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{z^4}\)>= \(\dfrac{1}{x.y.z}\)
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 \(x=y=z=1\)
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=z^2+\left(x+y\right)^2+2z\left(x+y\right)=36\)
áp dụng BĐT cosi :
\(z^2+\left(x+y\right)^2\ge2z\left(x+y\right)\)
<=> \(z^2+\left(x+y\right)^2+2z\left(x+y\right)\ge4z\left(x+y\right)=36< =>z\left(x+y\right)\ge9\)
ta lại có \(\dfrac{x+y}{xyz}=\dfrac{x}{xyz}+\dfrac{y}{xyz}=\dfrac{1}{yz}+\dfrac{1}{xz}\) áp dụng BĐT buhihacopxki dạng phân thức => \(\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{4}{yz+xz}=\dfrac{4}{z\left(x+y\right)}\ge\dfrac{4}{9}\left(đpcm\right)\)
dấu bằng xảy ra khi \(\left[{}\begin{matrix}yz=xz< =>x=y\\x+y+z=6\\z^2=\left(x+y\right)^2\end{matrix}\right.< =>\left[{}\begin{matrix}x+y+z=6\\z=2x=2y\end{matrix}\right.< =>\left[{}\begin{matrix}x=y=\dfrac{3}{2}\\z=3\end{matrix}\right.\)
-Ủa vì sao\(\dfrac{4}{z\left(x+y\right)}\ge\dfrac{4}{9}\)? Đáng lẽ là \(\dfrac{4}{z\left(x+y\right)}\le\dfrac{4}{9}\) chứ?
Áp dụng bđt phụ \(\dfrac{1}{A+B}\le\dfrac{1}{4}\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\forall A,B>0\)
\(\dfrac{1}{2x+y+z}=\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\) Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Rightarrow\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=1\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{4}\)
Này Nguyễn Trọng Chiến, mk ko hiểu cái chỗ \(\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)??? Sao suy ra được vậy bn??
\(\dfrac{1}{x+x+y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{4}\)
Mk ko hiểu cái dòng đầu Nguyễn Việt Lâm Giáo viên, bn có thể nói chi tiết cách phân tích cho mk đc ko??
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
Sửa đề: \(\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge\dfrac{3}{4}\)
Đặt \(P=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(P=\dfrac{x+1}{x+1}-\dfrac{1}{x+1}+\dfrac{y+1}{y+1}-\dfrac{1}{y+1}+\dfrac{z+1}{z+1}-\dfrac{1}{z+1}\)
\(P=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\)
\(P=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\)
Ta có:
\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{9}{x+y+z+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{9}{4}\) ( vì \(x+y+z=1\) )
\(\Rightarrow P\ge3-\dfrac{9}{4}=\dfrac{3}{4}\left(đpcm\right)\)
Dấu "=" xảy ra khi \(x+1=y+1=z+1\)
\(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Vậy \(Max_P=\dfrac{3}{4}\) khi \(x=y=z=\dfrac{1}{3}\)
