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Ta có \(\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{1}{1+2z}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{1+2x}=1-\frac{1}{1+2y}+1-\frac{1}{1+2z}\\\frac{1}{1+2y}=1-\frac{1}{1+2x}+1-\frac{1}{1+2y}\\\frac{1}{1+2z}=1-\frac{1}{1+2x}+1-\frac{1}{1+2y}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{1+2x}=\frac{2y}{1+2y}+\frac{2z}{1+2z}\\\frac{1}{1+2y}=\frac{2x}{1+2x}+\frac{2y}{1+2y}\\\frac{1}{1+2z}=\frac{2x}{1+2x}+\frac{2y}{1+2y}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{1+2x}=\frac{2y}{1+2y}+\frac{2z}{1+2z}\ge2\sqrt{\frac{4yz}{\left(1+2y\right)\left(1+2z\right)}}\\\frac{1}{1+2y}=\frac{2x}{1+2x}+\frac{2z}{1+2z}\ge2\sqrt{\frac{4xz}{\left(1+2x\right)\left(1+2z\right)}}\\\frac{1}{1+2z}=\frac{2x}{1+2x}+\frac{2y}{1+2y}\ge2\sqrt{\frac{4xy}{\left(1+2x\right)\left(1+2y\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\ge8\sqrt{\frac{64x^2y^2z^2}{\left(1+2x\right)^2\left(1+2y\right)^2\left(1+2x\right)^2}}\)
\(\Rightarrow\frac{1}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\ge\frac{64xyz}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\)
\(\Rightarrow1\ge64xyz\)
\(\Rightarrow xyz\le\frac{1}{64}\)( đpcm )
Dấu ' = ' xảy ra khi \(x=y=z=\frac{1}{4}\)
\(\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\)
= \(\frac{1}{x^2+y^2+y^2+1+2}+\frac{1}{y^2+z^2+z^2+1+2}+\frac{1}{z^2+x^2+x^2+1+2}\)
\(\le\frac{1}{2xy+2y+2}+\frac{1}{2yz+2z+2}+\frac{1}{2zx+2x+2}\)
= \(\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{xyzx+yzx+zx}+\frac{x}{yzx+zx+x}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{x+1+zx}+\frac{x}{1+zx+x}+\frac{1}{zx+x+1}\right)\)
= 1/2
Dấu "=" xảy ra <=> x = y =z =1
Áp dụng BĐT AM-GM ta có:\(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+1\ge2y\end{cases}\Rightarrow\frac{1}{x^2+2y^2+3}\le\frac{1}{2xy+2y+2}}\)
Tương tự ta cũng có
\(\frac{1}{y^2+2x^2+3}\le\frac{1}{2yz+2z+2};\frac{1}{z^2+2x^2+3}\le\frac{1}{2xz+2x+2}\)
Do đó ta có:\(VT\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
Mặt khác, do xyz=1 nên ta có:
\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{1}{xy+y+1}+\frac{y}{xy+y+1}+\frac{xy}{xy+y+1}\)
\(=\frac{xy+y+1}{xy+y+1}=1\)
\(\Rightarrow VT\le\frac{1}{2}\). Dấu "=" xảy ra <=> x=y=z=1
\(\Sigma\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\Sigma\left(\dfrac{1}{9}.\dfrac{a^2\left(2+1\right)^2}{2a.\left(\Sigma a\right)+2a^2+bc}\right)\le\Sigma\left(\dfrac{1}{9}.\dfrac{4a^2}{2a\left(\Sigma a\right)}+\dfrac{1}{9}.\dfrac{a^2}{2a^2+bc}\right)\)
\(=\Sigma\left(\dfrac{1}{9}.\left(\dfrac{2a}{\Sigma a}+\dfrac{a^2}{2a^2+bc}\right)\right)=\dfrac{1}{9}\left(2+\Sigma\dfrac{a^2}{2a^2+bc}\right)\)
Cần chứng minh \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
<=> \(\Sigma\frac{bc}{2a^2+bc}\ge1\) (*)
Đặt (x;y;z) -------> \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Suy ra (*) <=> \(\Sigma\frac{x^2}{x^2+2xy}\ge1\Leftrightarrow\frac{\Sigma x^2}{\Sigma x^2}\ge1\) (đúng)
Vậy \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
Suy ra \(\Sigma\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\le\frac{1}{9}\left(2+\Sigma\frac{a^2}{2a^2+bc}\right)\le\frac{1}{9}\left(2+1\right)=\frac{1}{3}\)
Đẳng thức xảy ra <=> x = y = z = 1
HSG toán 9 Quảng Nam năm 2018-2019
Giải: Từ đẳng thức đã cho suy ra: \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\). Áp dụng (a+b)2 >= 4ab ta có:
\(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\cdot\left(\frac{2x+y}{2}\right)\cdot\frac{3y}{2}\)
\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\). Dấu "=" xảy ra <=> x=y
\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự \(\hept{\begin{cases}\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\\\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\end{cases}}\)
\(\Rightarrow A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\left("="\Leftrightarrow x=y=z\right)\)
Ta có \(\sqrt{\left(2x-1\right)\cdot1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{2x-1}\le2\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}},\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)Do đó:
\(A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)
Dấu "=" xảy ra <=> x=y=z=1
Vậy GTLN của A=3 đạt được khi x=y=z=1
Có BĐT phụ:
\(a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
Áp dụng
\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\)
\(\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{zx\left(z+x\right)+xyz}\)
\(=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}\)
\(=\frac{1}{xyz}\)
Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)ta có :
\(\frac{16}{3x+3y+2z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)
\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)
\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)
Cộng theo vế 3 đẳng thức trên ta được :
\(16.\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\le4.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4.6=24\)
\(\Rightarrow\)\(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)
Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath