Cho \(x,y,z\) là ba số dương . Chứng minh rằng
\(x^3+y^3+z^3\ge x^2y+y^2z+z^2x\)
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NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 12 2021
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
17 tháng 7 2021
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
LH
1
G
9 tháng 3 2019
Áp dụng bđt Cauchy-Schwarz:
\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)
\(=\frac{x}{\left(x+y\right)+\left(x+z\right)}+\frac{y}{\left(x+y\right)+\left(y+z\right)}+\frac{z}{\left(y+z\right)+\left(x+z\right)}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{4}\)
\("="\Leftrightarrow x=y=z\)
20 tháng 4 2015
đặt a = 2x + y + z; b = 2y + z + x; c = 2z + x + y (a; b ; c > 0)
=> a + b + c = 4.(x+ y + z) => x + y + z = (a+ b+ c) / 4
=> x = a - (x+ y + z) = a - (a+ b + c) / 4
y = b - (x + y + z) = b - (a+b+c) / 4
z = c - (x+y + z) = c - (a+b+c)/ 4
Khi đó : \(VT=1-\frac{a+b+c}{4a}+1-\frac{a+b+c}{4b}+1-\frac{a+b+c}{4c}\)
\(VT=3-\left(\frac{a+b+c}{4a}+\frac{a+b+c}{4b}+\frac{a+b+c}{4c}\right)=3-\frac{1}{4}.\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(VT=3-\frac{1}{4}.\left(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\right)=3-\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\)
Với a, b > 0 ta có: a/b + b/ a > = 2
=> \(\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\ge\frac{1}{4}.\left(3+2+2+2\right)=\frac{9}{4}\)
=> \(VT\le3-\frac{9}{4}=\frac{3}{4}\)
Dấu = xảy ra khi a= b = c => x = y = z
TN
27 tháng 5 2018
\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)
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
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)
\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\)
NH
0
KK
9 tháng 2 2017
\(\frac{x^3}{2x+3y+5z}+\frac{y^3}{2y+3z+5x}+\frac{z^3}{2z+3x+5y}\)
\(\Leftrightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3zy+5xy}+\frac{z^4}{2z^2+3xz+5yz}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2x^2+2y^2+2z^2+8xy+8yz+8xz}\)
\(\Leftrightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
Xét \(\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}x^2+y^2\ge2\sqrt{x^2y^2}=2xy\\y^2+z^2\ge2\sqrt{y^2z^2}=2yz\\x^2+z^2\ge2\sqrt{x^2z^2}=2xz\end{matrix}\right.\)
Cộng từng vế:
\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
\(\Rightarrow xy+yz+xz\le x^2+y^2+z^2\)
\(\Rightarrow8\left(xy+yz+xz\right)\le8\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(\Rightarrow\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{10}\)
Ta có: \(x^2+y^2+z^2\ge\frac{1}{3}\)
\(\Rightarrow\frac{x^2+y^2+z^2}{10}\ge\frac{1}{30}\)
\(\Rightarrow\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\ge\frac{1}{30}\)
Vì \(\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
\(\Rightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{1}{30}\)
\(\Leftrightarrow\frac{x^3}{2x+3y+5z}+\frac{y^3}{2y+3z+5x}+\frac{z^3}{2z+3x+5y}\ge\frac{1}{30}\) ( đpcm )
chịu chết
Áp dụng bất đẳng thức AM-GM ta có :
\(x^3+x^3+y^3\ge3\sqrt[3]{x^6y^3}=3x^2y\)(1)
\(y^3+y^3+z^3\ge3\sqrt[3]{y^6z^3}=3y^2z\)(2)
\(z^3+z^3+x^3\ge3\sqrt[3]{z^6x^3}=3z^2x\)(3)
Cộng (1),(2),(3) theo vế
=> \(x^3+x^3+y^3+y^3+y^3+z^3+z^3+z^3+x^3\ge3x^2y+3y^2z+3z^2x\)
<=> \(3\left(x^3+y^3+z^3\right)\ge3\left(x^2y+y^2z+z^2x\right)\)
<=> \(x^3+y^3+z^3\ge x^2y+y^2z+z^2x\left(đpcm\right)\)
Đẳng thức xảy ra <=> x=y=z
p dụng bất đẳng thức Cô si cho ba số dương x^3,x^3,y^3x3,x3,y3.