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NN
1
25 tháng 7 2018
Đặt \(\hept{\begin{cases}2x+y+z=4a\\2y+x+z=4b\\2z+x+y=4c\end{cases}\Rightarrow}\hept{\begin{cases}x=3a-b-c\\y=3b-c-a\\z=3c-a-b\end{cases}}\)thay vào biểu thức đó
\(\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{3}{4}-\frac{b-c}{4a}+\frac{3}{4}-\frac{c-a}{4b}+\frac{3}{4}-\frac{a-b}{4c}\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
Áp dụng BĐT sau: \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\ge6\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\ge\frac{6}{4}\)
\(\Leftrightarrow\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\le\frac{3}{4}\)
Từ đó ta có: \(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{3}{4}\)(đpcm).
Dấu "=" xảy ra <=> x=y=z.
13 tháng 6 2018
Bài của lớp 7 ghê vậy!!
Áp dụng bất đẳng thức Cauchy cho 3 số dương x,y,z
ta có bổ đề \((a+b+c)({1\over a}+{1\over b}+{1\over c})\) > 9
Áp dụng vào ta có
\(D*({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z})\) >9(1)
Ta có \({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z}\) =\(2+{y+z\over x}+2+{z+x\over y}+2+{y+x\over z}\)=\(6-3+{y+z\over x}+1+{z+x\over y}+1+{y+x\over z}+1\)=\(3+{x+y+z\over x}+{y+x+z\over y}+{z+y+x\over z}\)=\(3+(x+y+z)({1\over x}+{1\over y}+{1\over z})\) > 3+9=12
thay vào(1)
Ta có \(D \) < \({9\over 12}\)=\({3\over 4}\)
Dấu "=" xảy ra khi x=y=z
=> ĐPCM
YY
13 tháng 6 2018
áp dụng bất đẳng thức phụ : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\frac{x}{2x+y+z}=\frac{x}{x+y+x+z}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
\(\frac{y}{2y+x+z}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right)\)
\(\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)
cộng vế theo vế
\(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{1}{4}\cdot3=\frac{3}{4}\)(đpcm)
23 tháng 3 2020
Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)
\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\)
\( \implies\) \(a+b+c=4x+4y+4z\)
\( \implies\) \(x+y+z=\frac{a+b+c}{4}\)
+)Ta có : \(a=2x+y+z\)
\(\iff\) \(a=x+\left(x+y+z\right)\)
\(\iff\) \(a-\left(x+y+z\right)=x\)
\(\iff\) \(a-\frac{a+b+c}{4}=x\)
\(\iff\) \(x=\frac{3a-b-c}{4}\)
+)Ta có :\(b=2y+z+x\)
\(\iff\) \(b=y+\left(y+z+x\right)\)
\(\iff\)\(b-\left(y+z+x\right)=y\)
\(\iff\) \(b-\frac{a+b+c}{4}=y\)
\(\iff\)\(y=\frac{3b-c-a}{4}\)
+)Ta có :\(c=2z+x+y\)
\(\iff\) \(c=z+\left(z+x+y\right)\)
\(\iff\) \(c-\left(z+x+y\right)=z\)
\(\iff\) \(c-\frac{a+b+c}{4}=z\)
\(\iff\)\(z=\frac{3c-a-b}{4}\)
\( \implies\) \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)
Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được :
\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2
\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2
\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 6
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)
\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\)
20 tháng 7 2017
Cậu vào đây nha !
Câu hỏi của doanthihuong - Toán lớp 7 - Học toán với OnlineMath
LH
1
G
9 tháng 3 2019
Áp dụng bđt Cauchy-Schwarz:
\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)
\(=\frac{x}{\left(x+y\right)+\left(x+z\right)}+\frac{y}{\left(x+y\right)+\left(y+z\right)}+\frac{z}{\left(y+z\right)+\left(x+z\right)}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{4}\)
\("="\Leftrightarrow x=y=z\)
AH
Akai Haruma
Giáo viên
27 tháng 8 2023
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 $x^2y^2z^2=1$
LD
14 tháng 1 2019
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
Theo Cauchy Schwarz:
\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{2y+z+x}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right);\frac{z}{2z+y+x}\le\frac{1}{4}\left(\frac{z}{z+y}+\frac{z}{z+x}\right)\)
Cộng lại:
\(D\le\frac{3}{4}\left(đpcm\right)\)