Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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)
\(-\text{Theo bài ra: }D=\dfrac{x}{2x+y+z}+\dfrac{y}{2y+z+x}+\dfrac{z}{2z+x+y}\)
\(-\text{Đặt }\left\{{}\begin{matrix}a=2x+y+z\\b=2y+z+x\\c=2z+x+y\end{matrix}\right.\Rightarrow a+b+c=4\left(x+y+z\right)\)
\(\Rightarrow a-\dfrac{a+b+c}{4}=x\)
\(\Rightarrow x=\dfrac{3a-b-c}{4}\)
\(-\text{Tương tự: }\left\{{}\begin{matrix}y=\dfrac{3b-c-a}{4}\\z=\dfrac{3c-a-b}{4}\end{matrix}\right.\)
Suy ra \(D=\dfrac{3a-b-c}{4a}+\dfrac{3b-3c-a}{4b}+\dfrac{3c-a-b}{4c}\)
\(D=\dfrac{9}{4}-\left(\dfrac{b}{4a}+\dfrac{c}{4a}+\dfrac{c}{4b}+\dfrac{a}{4b}+\dfrac{a}{4c}+\dfrac{b}{4c}\right)\)
\(D=\dfrac{9}{4}-\dfrac{1}{4}\left[\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\right]\)
- Theo bất đẳng thức Cosi, ta có: \(\left\{{}\begin{matrix}\dfrac{b}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{b}+\dfrac{b}{c}\ge2\end{matrix}\right.\)
Suy ra \( D\le\dfrac{9}{4}-\dfrac{1}{4}.6=\dfrac{9}{4}-\dfrac{6}{4}=\dfrac{3}{4}\)
Vậy \(D\le\dfrac{3}{4}\left(đpcm\right)\)
Cậu vào đây nha !
Câu hỏi của doanthihuong - Toán lớp 7 - Học toán với OnlineMath
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
\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
\(\frac{y}{2y+x+z}=\frac{y}{\left(x+y\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\)
\(\frac{z}{2z+x+y}=\frac{z}{\left(x+z\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\)
Cộng theo vế:
\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{x}{x+z}+\frac{z}{x+z}\right)=\frac{3}{4}\)
Đặt \(\hept{\begin{cases}2x+y+z=a\\2y+z+x=b\\2z+x+y=c\end{cases}}\Rightarrow a+b+c=4\left(x+y+z\right)=\)
\(4\left(a-x\right)=4\left(b-y\right)=4\left(c-z\right)\Rightarrow\hept{\begin{cases}4x=3a-b-c\\4y=3b-c-a\\4z=3c-a-b\end{cases}}\)
Lúc đó thì \(4VT=\frac{3a-b-c}{a}+\frac{3b-c-a}{b}+\frac{3c-a-b}{c}\)
\(=3-\frac{b}{a}-\frac{c}{a}+3-\frac{c}{b}-\frac{a}{b}+3-\frac{a}{c}-\frac{b}{c}\)
\(=9-\left(\frac{a}{b}+\frac{b}{a}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{c}{a}+\frac{a}{c}\right)\le3\)
\(\Rightarrow VT\le\frac{3}{4}\)
Đẳng thức xảy ra khi a = b = c hay x = y = z