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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 2 2020
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)
\(P=a^2-2ab+3b^2-2a+2009,5\)
\(P=\frac{1}{3}\left(9b^2-6ab+a^2\right)+\frac{2}{3}\left(a^2-3a+\frac{9}{4}\right)+2008\)
\(P=\frac{1}{3}\left(3b-a\right)^2+\frac{2}{3}\left(a-\frac{3}{2}\right)^2+2008\ge2008\)
\(P_{min}=2008\) khi \(\left\{{}\begin{matrix}a-\frac{3}{2}=0\\3b-a=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{2}\\b=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{9}{4}\\y=\frac{1}{4}\end{matrix}\right.\)
10 tháng 3 2020
Thay \(xy+yz+zx=5\) vào P, ta có:
\(P=\frac{3x+3y+2z}{\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}}\)
Áp dụng bất đẳng thức Cô-si, ta có:
\(\sqrt{6\left(x+y\right)\left(x+z\right)}\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}\)
\(\sqrt{6\left(y+z\right)\left(y+x\right)}\le\frac{3\left(y+x\right)+2\left(y+z\right)}{2}\)
\(\sqrt{\left(z+x\right)\left(z+y\right)}\le\frac{\left(z+x\right)+\left(z+y\right)}{2}\)
Cộng vế theo vế các bất đẳng thức cùng chiều, ta đươc:
\(\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}\le\frac{9}{2}x+\frac{9}{2}y+3z\)
\(\Rightarrow P\ge\frac{3x+3y+2z}{\frac{9}{2}x+\frac{9}{2}y+3z}=\frac{3x+3y+2z}{\frac{3}{2}\left(3x+3y+2z\right)}=\frac{2}{3}\)
Dấu "=" khi \(\hept{\begin{cases}3\left(x+y\right)=2\left(y+z\right)=2\left(z+x\right)\\z+y=z+x\\xy+yz+zx=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}}\)
5 tháng 2 2020
Câu 1:
\(y^2+yz+z^2=1-\frac{3x^2}{2}\)
\(\Leftrightarrow2y^2+2yz+2z^2=2-3x^2\)
\(\Leftrightarrow\left(y+z\right)^2+y^2+z^2+3x^2=2\)
\(\Leftrightarrow\left(y+z\right)^2+x^2+2x\left(y+z\right)+y^2+z^2+2x^2-2x\left(y+z\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2=2-\left(x-y\right)^2-\left(x-z\right)^2\)
\(\Leftrightarrow A^2=2-\left[\left(x-y\right)^2+\left(x-z\right)^2\right]\le2\forall x;y;z\)
\(\Leftrightarrow-\sqrt{2}\le A\le\sqrt{2}\)
Vậy \(A_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=y=z\\x+y+z=-\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow x=y=z=\frac{-\sqrt{2}}{3}\)
\(A_{max}=\sqrt{2}\Leftrightarrow a=b=c=\frac{\sqrt{2}}{3}\)
Câu 2:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\ge\frac{9}{3+x^2+y^2+z^2}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Câu 3:
\(P=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\) ( \(a\ge3;b\ge4;c\ge2\) )
\(P=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng BĐT Cauchy:
\(P=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}\cdot\sqrt{c-2}}{c}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}\cdot\sqrt{a-3}}{a}+\frac{1}{2}\cdot\frac{2\cdot\sqrt{b-4}}{b}\)
\(\le\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\cdot\frac{2+c-2}{c}+\frac{1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\frac{3+a-3}{a}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{4+b-4}{b}=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{2}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)
Câu 4:
Đặt \(\sqrt{x}=a;\sqrt{y}=b\left(a;b\ge0\right)\)
\(M=a^2-2ab+3b^2-2a+1\)
\(M=a^2-a\left(2b+2\right)+3b^2+1\)
\(\Delta=\left(2b+2\right)^2-4\left(3b^2+1\right)\)
\(=-8b^2+8b\)
\(=-8b\left(b+1\right)\ge0\)
Vì \(b\ge0\) nên \(-8b\left(b+1\right)\le0\)
Dấu "=" xảy ra \(\Leftrightarrow b=0\)
Khi đó \(M=a^2-2a+1=\left(a-1\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow a=1\)
Vậy \(M_{min}=1\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
13 tháng 10 2019
\(5\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)\(\Leftrightarrow\)\(x+y+z\ge\sqrt{15}\)
\(\frac{x^2}{\sqrt{8x^2+3y^2+14xy}}=\frac{x^2}{\sqrt{8x^2+2xy+3y^2+12xy}}\ge\frac{x^2}{\sqrt{9x^2+12xy+4y^2}}=\frac{x^2}{3x+2y}\)
\(A\ge sigma\frac{x^2}{3x+2y}\ge\frac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=\frac{x+y+z}{5}\ge\sqrt{\frac{3}{5}}\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{\frac{5}{3}}\)
NB
1
TN
29 tháng 7 2017
Ta có: \(\sqrt{x^2+xy+y^2}=\sqrt{x^2+xy+\frac{y^2}{4}+\frac{3y^2}{4}}=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}}\)
Tương tự ta viết lại A và áp dụng BĐT Mipcopxki :
\(A=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}}+\sqrt{\left(y+\frac{z}{2}\right)^2+\frac{3z^2}{4}}+\sqrt{\left(z+\frac{x}{2}\right)^2+\frac{3x^2}{4}}\)
\(=\sqrt{\left(x+\frac{y}{2}\right)^2+\left(\frac{\sqrt{3}y}{2}\right)^2}+\sqrt{\left(y+\frac{z}{2}\right)^2+\left(\frac{\sqrt{3}z}{2}\right)^2}+\sqrt{\left(z+\frac{x}{2}\right)^2+\left(\frac{\sqrt{3}x}{2}\right)^2}\)
\(\ge\sqrt{\left(\frac{3\left(x+y+z\right)}{2}\right)^2+\left(\frac{\sqrt{3}\left(x+y+z\right)}{2}\right)^2}\)
\(\ge\sqrt{\left(\frac{3\cdot3}{2}\right)^2+\left(\frac{\sqrt{3}\cdot3}{2}\right)^2}=\sqrt{27}\)
Xảy ra khi x=y=z=1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 6 2020
\(x^2+5=x^2+xy+yz+zx=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P=\frac{3x+3y+2z}{\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}}\)
\(P=\frac{3x+3y+2z}{\sqrt{\left(3x+3y\right)\left(2x+2z\right)}+\sqrt{\left(3x+3y\right)\left(2y+2z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}}\)
\(P\ge\frac{2\left(3x+3y+2z\right)}{3x+3y+2x+2z+3x+3y+2y+2z+x+z+y+z}\)
\(P\ge\frac{2\left(3x+3y+2z\right)}{9x+9y+6z}=\frac{2\left(3x+3y+2z\right)}{3\left(3x+3y+2z\right)}=\frac{2}{3}\)
\(P_{min}=\frac{2}{3}\) khi \(\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
\(2P=2x-4\sqrt{xy}+6y-4\sqrt{x}+4019\)
\(=\left(\left(x-4\sqrt{xy}+y\right)-\frac{2}{2}.\left(\sqrt{x}-2\sqrt{y}\right)+\frac{1}{4}\right)+\left(x-\frac{2.3.\sqrt{x}}{2}+\frac{9}{4}\right)+2\left(y-\frac{2\sqrt{y}}{2}+\frac{1}{4}\right)+4016\)
\(=\left(\left(\sqrt{x}-2\sqrt{y}\right)^2-\frac{2}{2}.\left(\sqrt{x}-2\sqrt{y}\right)+\frac{1}{4}\right)+\left(x-\frac{2.3.\sqrt{x}}{2}+\frac{9}{4}\right)+2\left(y-\frac{2\sqrt{y}}{2}+\frac{1}{4}\right)+4016\)
\(=\left(\sqrt{x}-2\sqrt{y}-\frac{1}{2}\right)^2+\left(\sqrt{x}-\frac{3}{2}\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2+4016\ge2016\)
\(\Rightarrow P\ge2008\)khi \(\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{1}{4}\end{cases}}\)
tung hỏa mù hả sao tăng Hệ số lên làm gì?
căn x=a, căn y=b
P=(a^2+b^2-2ab-2a+2b+1)+(2b^2-2b+1/2)+2009+1/2-(1+1/2)
P=(a-b-1)^2+2(b-1/2)^2+2008>=2008
đăng thức b=1/2=>y=1/4; và a-1/2-1=0=>a=3/2=>x=9/4