Cho (x+\(\sqrt{y^2+1}\))(y+\(\sqrt{x^2+1}\))=1
Tìm GTNN của P=2(x2+y2)+x+y
K
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27 tháng 1 2022
Ta có: \(\sqrt{\left(x^2+\dfrac{1}{y^2}\right)\left(1+81\right)}\ge\sqrt{\left(x+\dfrac{9}{y}\right)^2}\)
=> \(\sqrt{x^2+\dfrac{1}{y^2}}\ge\dfrac{x+\dfrac{9}{y}}{\sqrt{82}}\)
Tương tự => \(\left\{{}\begin{matrix}\sqrt{y^2+\dfrac{1}{z^2}}\ge\dfrac{y+\dfrac{9}{z}}{\sqrt{82}}\\\sqrt{z^2+\dfrac{1}{x^2}}\ge\dfrac{z+\dfrac{9}{x}}{\sqrt{82}}\end{matrix}\right.\)
=> \(P\ge\dfrac{\left(x+y+z\right)+9\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}{\sqrt{82}}\)
Mà x + y + z = 1
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\)
=> \(P\ge\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
T
8 tháng 8 2021
Sử dụng bất đẳng thức Minkovski, ta có:
\(P = \sqrt {{{\left( {x + y + z} \right)}^2} + {{\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)}^2}} \)
\( \ge \sqrt {\left[ {{{\left( {x + y + z} \right)}^2} + \frac{1}{{{{\left( {x + y + z} \right)}^2}}}} \right] + \frac{{80}}{{{{\left( {x + y + z} \right)}^2}}}} \)
\(\ge \sqrt{2+\dfrac{80}{1}} =\sqrt{82}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}.\)
Kết luận ...
8 tháng 8 2021
\(\sqrt{x^2+\dfrac{1}{x^2}}=\dfrac{1}{\sqrt{82}}\sqrt{\left(1^2+9^2\right)\left(x^2+\dfrac{1}{x^2}\right)}\ge\dfrac{1}{\sqrt{82}}\left(x+\dfrac{9}{x}\right)\)
tương tự với \(\sqrt{y^2+\dfrac{1}{y^2}};\sqrt{z^2+\dfrac{1}{z^2}}\)
\(=>P\ge\dfrac{1}{\sqrt{81}}\left(x+\dfrac{9}{x}+y+\dfrac{9}{y}+z+\dfrac{9}{z}\right)\)
có \(x+\dfrac{9}{x}=x+\dfrac{1}{9x}+\dfrac{80}{9x}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{80}{9x}\)
tương tự với \(y+\dfrac{9}{y};z+\dfrac{9}{z}\)
\(=>P\ge\dfrac{1}{\sqrt{82}}\left[2\sqrt{\dfrac{1}{9}}.3+\dfrac{\left(\sqrt{80}+\sqrt{80}+\sqrt{80}\right)^2}{9\left(x+y+z\right)}\right]=\dfrac{1}{\sqrt{82}}.82=\sqrt{82}\)
dấu"=" xảy ra<=>x=y=z=1/3
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)
\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)
\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)
Đồng thời: \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^2=\dfrac{2}{9}\)
\(\Rightarrow-\left(x^2+y^2\right)\le-\dfrac{2}{9}\)
Từ đó ta có:
\(A=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1-\left(x+y\right)}{x+y}=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1}{x+y}-1\)
\(A\le\sqrt{2\left[2-\left(x^2+y^2\right)\right]}+\dfrac{1}{x+y}-1\le\sqrt{2\left(2-\dfrac{2}{9}\right)}+\dfrac{3}{2}-1=\dfrac{3+8\sqrt{2}}{6}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 11 2021
\(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2+x+1\le x^2+2x+1\\2y^2+y+1\le y^2+2y+1\\2z^2+z+1\le z^2+2z+1\end{matrix}\right.\)
\(\Rightarrow P\le\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}=x+y+z+3=4\)
\(P_{max}=4\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
ND
0
Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\left(\dfrac{b^2-1}{2b}\right)+1}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\left(\dfrac{a^2-1}{2a}\right)+1}\right)=1\)
\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)
\(\Rightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4\left(ab\right)^2}+\dfrac{\left(a-b\right)^2}{4ab}=0\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)
\(\Rightarrow\left(1-\dfrac{1}{ab}\right)\left(\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right)=0\)
\(\Rightarrow1-\dfrac{1}{ab}=0\Rightarrow ab=1\)
\(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Rightarrow x+y=0\Rightarrow y=-x\)
\(P=2\left(x^2+\left(-x\right)^2\right)+0=4x^2\ge0\)
Dấu "=" xảy ra khi \(x=y=0\)