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Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\dfrac{x}{2x+y+z}\)=\(\dfrac{y}{2y+x+z}\)=\(\dfrac{z}{2z+x+y}\)=\(\dfrac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}\)=\(\dfrac{x+y+z}{3x+3y+3z}\)=\(\dfrac{x+y+z}{3.\left(x+y+z\right)}\)=\(\dfrac{1}{3}\)=\(\dfrac{3}{9}\)<\(\dfrac{3}{4}\)(đpcm)
Câu b mình vừa làm rồi
a)
Áp dụng bđt Cauchy-Scharz:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\)
\(=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(x+y\right)+\left(y+z\right)}+\dfrac{z}{\left(x+z\right)+\left(y+z\right)}\)
\(\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)
\(=\dfrac{1}{4}.3=\dfrac{3}{4}\)
Dấu "=" khi \(x=y=z\)
Em ko nhớ là lớp 7 có học Cô-si nên chị đừng giải theo cách đó
Lời giải:
Từ đkđb suy ra:
$x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}$
$y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}$
$z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}$
$\Rightarrow (x-y)(y-z)(z-x)=\frac{(y-z)(z-x)(x-y)}{(xyz)^2}$
$\Leftrightarrow (x-y)(y-z)(z-x)(1-\frac{1}{x^2y^2z^2})=0$
$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $1-\frac{1}{x^2y^2z^2}=1$
$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $x^2y^2z^2=1$
Nếu $(x-y)(y-z)(z-x)=0$
$\Rightarrow x=y$ hoặc $y=z$ hoặc $z=x$
Không mất tquat giả sử $x=y$. Khi đó: $\frac{1}{y}=\frac{1}{z}$
$\Rightarrow y=z$
$\Rightarrow x=y=z$. Tương tự khi xét $y=z$ hoặc $z=x$ thì ta cũng thu được $x=y=z$
Vậy $x=y=z$ hoặc $x^2y^2z^2=1$
\(TH_1:x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\\ \Rightarrow Q=\dfrac{-z}{z}+\dfrac{-x}{x}+\dfrac{-y}{y}=-3\\ TH_2:x+y+z\ne0\\ \Rightarrow\dfrac{3x-2y+z}{x}=\dfrac{3y-2z+x}{y}=\dfrac{3z-2x+y}{z}=\dfrac{2x+2y+2z}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}3x-2y+z=x\\3y-2z+x=y\\3z-2x+y=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-2y=-z\\2y-2z=-x\\2z-2x=-y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{z}{2}\\y-z=-\dfrac{x}{2}\\z-x=-\dfrac{y}{2}\end{matrix}\right.\)
\(\Rightarrow Q=-\dfrac{z}{2}:z-\dfrac{x}{2}:x-\dfrac{y}{2}:y=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
Đề nhảm.a;b;c ở đâu bạn -_-
a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel:
\(\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\\\dfrac{y}{2y+x+z}=\dfrac{y}{x+y+y+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)\\\dfrac{z}{2z+x+y}=\dfrac{z}{x+z+y+z}\le\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z>0\)
b) Áp dụng bất đẳng thức AM-GM:
\(\left\{{}\begin{matrix}\left(a+b-c\right)\left(a-b+c\right)\le\dfrac{\left(a+b-c+a-b+c\right)^2}{4}=\dfrac{4a^2}{4}=a^2\\\left(a-b+c\right)\left(-a+b+c\right)\le\dfrac{\left(a-b+c-a+b+c\right)^2}{4}=\dfrac{4c^2}{4}=c^2\\\left(a+b-c\right)\left(-a+b+c\right)\le\dfrac{\left(a+b-c-a+b+c\right)^2}{4}=\dfrac{4b^2}{4}=b^2\end{matrix}\right.\)
Nhân theo vế: \(\left[\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\right]^2\le\left(abc\right)^2\)
\(\Rightarrow\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\le abc\)
Dấu "=" xảy ra khi: \(a=b=c>0\)
Phải chứng minh BĐT trung gian: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\forall\) a,b trước khi áp dụng chứ.
TH1: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(y+z=-x\)
\(x+z=-y\)
\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)
TH2: \(x+y+z\ne0\)
\(\Rightarrow2x+2y-z=3\)
\(\Rightarrow2x+2y=4z\)
\(\Rightarrow x+y=2z\)
\(x+z=2y\)
\(y+z=2x\)
\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)
Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)
Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)
Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)
\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)
Vậy \(M=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-y+2z}{y}=\dfrac{-x+2y+2z}{x}=\dfrac{2x+2y-z+2x-y+2z-x+2y+2x}{x+y+z}=\dfrac{3x+3y+3z}{x+y+z}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)
\(\Rightarrow\)\(\dfrac{2x+2y-z}{z}=3\Leftrightarrow2x+2y-z=3z\Leftrightarrow2\left(x+y\right)=4z\Leftrightarrow x+y=2z\Leftrightarrow z=\dfrac{x+y}{2}\)
Tương tự: \(x=\dfrac{y+z}{2}\)
\(y=\dfrac{x+z}{2}\)
Thay vào M, ta được:
\(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\left(\dfrac{y+z}{2}.\dfrac{x+z}{2}.\dfrac{x+y}{2}\right).8}\)
\(=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8}.8}=1\)
TH1 : \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)
Th2 : \(x+y+z\ne0\)
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)
\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)
\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)
\(\Leftrightarrow x=y=z\)
\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)
Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
cho hỏi VT là gì?