Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)
Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)
\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
tương tự với hai ông còn lại sau đó cộng lại ta được:
\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\geq \frac{16}{3x+3y+2z}\)
\(\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\geq \frac{16}{3x+2y+3z}\)
\(\frac{1}{z+y}+\frac{1}{z+y}+\frac{1}{x+z}+\frac{1}{x+y}\geq \frac{16}{2x+3y+3z}\)
Cộng theo vế:
\(\Rightarrow 4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\geq 16\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\Rightarrow \frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\leq \frac{4.6}{16}=\frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)
CMTT và cộng theo vế:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)
Trai Vô Đối câu này đề thi vô lớp 10 tỉnh Thanh Hóa ( tất cả thí sinh nek .... lúc nào rảnh mik đăng lên thử xem sao )
Lời giải:
Áp dụng BĐT Bunhiacopxky ta có:
\(\left (\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)(x+x+x+y+y+z)\geq (1+1+1+1+1+1)^2\)
\(\Leftrightarrow \frac{3}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{36}{3x+2y+z}\)
Thực hiện tương tự:
\(\frac{3}{y}+\frac{2}{z}+\frac{1}{x}\geq \frac{36}{3y+2z+x}\)
\(\frac{3}{z}+\frac{2}{x}+\frac{1}{y}\geq \frac{36}{3z+2x+y}\)
Cộng theo vế các BĐT vừa có thu được:
\(6\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 36\left(\frac{1}{3x+2y+z}+\frac{1}{3y+2z+x}+\frac{1}{3z+2x+y}\right)\)
\(\Leftrightarrow 72\geq 36\left(\frac{1}{3x+2y+z}+\frac{1}{3y+2z+x}+\frac{1}{3z+2x+y}\right)\)
\(\Leftrightarrow P\leq 2\)
Vậy \(P_{\max}=2\). Dấu bằng xảy ra khi \(x=y=z=\frac{1}{4}\)
Áp dụng Bđt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Ta có:
\(\frac{1}{2x+3y+3z}=\frac{1}{\left(x+2y+z\right)+\left(x+y+2z\right)}\)\(\le\frac{1}{4}\left(\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=\frac{1}{4}\cdot\left(\frac{1}{\left(x+y\right)+\left(y+z\right)}+\frac{1}{x+z}+\frac{1}{z+y}\right)\)
\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\right]+\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(=\frac{1}{16}\left(6+\frac{1}{y+z}\right)\).Tương tự với 2 cái còn lại r` cộng lại ta đc:
\(P\le\frac{1}{16}\left[6+6+6+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right]=\frac{3}{2}\)
a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)
Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)
=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)
Lời giải:
Biến đổi: \(P=\frac{3}{4}-\frac{y+z}{4x}+\frac{3}{4}-\frac{x+z}{4y}+\frac{3}{4}-\frac{x+y}{4z}\)
\(\Leftrightarrow P=\frac{9}{4}-\frac{1}{4}\underbrace{\left(\frac{x+y}{z}+\frac{x+z}{y}+\frac{z+y}{x}\right)}_{M}\)
Xét M
Áp dụng BĐT Am-Gm: \(M\geq 3\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{xyz}}\)
Tiếp tục Am-Gm: \((x+y)(y+z)(z+x)\geq 2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8xyz\)
\(\Rightarrow M\geq 3\sqrt[3]{8}=6\)
Do đó \(P=\frac{9}{4}-\frac{M}{4}\leq \frac{9}{4}-\frac{6}{4}\Leftrightarrow M\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow x=y=z\)