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Theo đề bài ta có
\(1=x+y\ge2\sqrt{xy}\)
\(\Leftrightarrow xy\le\frac{1}{4}\)
\(A=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{x}\right)^2\)
\(=x^2+y^2+\frac{2y}{x}+\frac{2x}{y}+\frac{1}{x^2}+\frac{1}{y^2}\)
\(=\left(x^2+\frac{1}{16x^2}\right)+\left(y^2+\frac{1}{16y^2}\right)+2\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{15}{16}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\ge\frac{1}{2}+\frac{1}{2}+4+\frac{15}{16}.\frac{2}{xy}\)
\(\ge5+\frac{15}{16}.\frac{2}{\frac{1}{4}}=\frac{25}{2}\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
\(A\ge\frac{\left(x+y+z\right)^2}{3}+\frac{9}{x+y+z}=\frac{\left(x+y+z\right)^2}{3}+\frac{9}{8\left(x+y+z\right)}+\frac{9}{8\left(x+y+z\right)}+\frac{27}{4\left(x+y+z\right)}\)
\(A\ge3\sqrt[3]{\frac{81\left(x+y+z\right)^2}{3.64\left(x+y+z\right)\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{27}{4}\)
\(A_{min}=\frac{27}{4}\) khi \(x=y=z=\frac{1}{2}\)
Ta có : \(A=x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}+2\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(A=4+\frac{x^2+y^2}{x^2y^2}+\frac{2.\left(x^2+y^2\right)}{xy}=4+\frac{4}{x^2y^2}+\frac{8}{xy}\)
\(A=4\left(\frac{1}{xy}+1\right)^2\)
Mặt khác : \(xy\le\frac{x^2+y^2}{2}=2\Rightarrow\frac{1}{xy}\ge\frac{1}{2}\)
\(\Rightarrow A\ge4\left(\frac{1}{2}+1\right)^2=9\)
Vậy Min A = 9 khi x = y = \(\sqrt{2}\)
Ta có:
\(A=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+\left(y^2+\frac{1}{8y}+\frac{1}{8y}\right)+\left(z^2+\frac{1}{8z}+\frac{1}{8z}\right)+\frac{6}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\sqrt[3]{y^2.\frac{1}{8y}.\frac{1}{8y}}+3\sqrt[3]{z^2.\frac{1}{8z}.\frac{1}{8z}}+\frac{6}{8}\frac{9}{x+y+z}\)
\(=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{6}{8}.\frac{9}{\frac{3}{2}}=\frac{27}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/2
Vậy min A = 27/4 tại x = y = z = 1/2
1. x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)
2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)
3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)
áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)
\(1>=\left(x+y\right)^2>=\left(2\sqrt{xy}\right)^2=4xy\Rightarrow1>=4xy\Rightarrow\frac{1}{2}>=2xy\)(bđt cosi)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{1}{2}}\)
\(=\frac{4}{\left(x+y\right)^2}+2>=\frac{4}{1^2}+2=4+2=6\)
dấu = xảy ra khi \(x=y=\frac{1}{2}\)
vậy min \(\frac{1}{x^2+y^2}+\frac{1}{xy}=6\)khi \(x=y=\frac{1}{2}\)
a, Áp dụng BĐT cosi với ba số dương có:
\(\frac{1}{xy}+x+y\ge3\sqrt[3]{\frac{1}{xy}.x.y}=3\sqrt[3]{1}=3\)
=> \(\frac{1}{xy}\ge3-x-y=3-2=1\)
Dấu"=" xảy ra <=> x=y=1
Vậy min \(\frac{1}{xy}=1\) <=> x=y=1
b, Với x,y>0 .Áp dụng bđt svac-xơ có
\(\frac{1}{x}+\frac{1}{y}\ge\frac{\left(1+1\right)^2}{x+y}=\frac{4}{2}=2\)
Dấu "=" xảy ra <=> x=y=1
c,Có \(\frac{1}{xy}\ge1\) <=> \(1-xy\ge0\)
x2+y2=(x+y)2-2xy=4-2xy=2+2(1-xy) \(\ge2+2.0=2\)
Dấu"=" xảy ra <=> x=y=1
1.
Đầu tiên ta cm: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\forall a,b>0\)
Ta có:
\(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}\ge\frac{2\sqrt{ab}}{ab}=\frac{2}{\sqrt{ab}}\ge\frac{2}{\frac{a+b}{2}}=\frac{4}{a+b}\) (cô si)
Dấu "=" khi a = b.
Áp dụng:
\(\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy\) \(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)+\frac{5}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{\frac{1}{4xy}\cdot4xy}+\frac{5}{\left(x+y\right)^2}\)
\(=4+2+5=11\)
Vậy MinA = 11 khi \(x=y=\frac{1}{2}\)
\(P=\frac{x^2+1}{x^2-x+1}\Leftrightarrow x^2+1=P\left(x^2-x+1\right)\)
\(\Leftrightarrow x^2+1-Px^2+Px-P=0\)(*)
\(\Leftrightarrow\left(1-P\right)x^2+Px+\left(1-P\right)=0\)
\(\Delta=P^2-4\left(1-P\right)^2\)
\(=P^2-4\left(1-2P+P^2\right)=-3P^2+8P-4\)
Để P có GTNN và GTLN thì phương trình (*) có nghiệm
\(\Leftrightarrow\Delta\ge0\Leftrightarrow-3P^2+8P-4\ge0\)
\(\Leftrightarrow-3P^2+2P+6P-4\ge0\)
\(\Leftrightarrow-P\left(3P-2\right)+2\left(3P-2\right)\ge0\)
\(\Leftrightarrow\left(3P-2\right)\left(2-P\right)\ge0\)
\(\Leftrightarrow\frac{2}{3}\le P\le2\)
Vậy \(min_P=\frac{2}{3}\Leftrightarrow x=-1\); \(max_P=2\Leftrightarrow x=1\)
\(A=2+x+y+\frac{1}{x}+\frac{1}{y}+\frac{x}{y}+\frac{y}{x}\)
\(A=2+x+\frac{1}{2x}+y+\frac{1}{2y}+\frac{x}{y}+\frac{y}{x}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(A\ge2+2\sqrt{\frac{x}{2x}}+2\sqrt{\frac{y}{2y}}+2\sqrt{\frac{xy}{yx}}+\frac{4}{2\left(x+y\right)}=4+2\sqrt{2}+\frac{2}{x+y}\)
\(A\ge4+2\sqrt{2}+\frac{2}{\sqrt{2\left(x^2+y^2\right)}}=4+3\sqrt{2}\)
\(\Rightarrow A_{min}=4+3\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)
Áp dụng BĐT Cauchy-schwarz ta có:
\(8\ge x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow x+y\ge4\)
Dấu " = " xảy ra <=> x=y=2
Áp dụng BĐT Cauchy-schwarz ta có:
\(A\ge\frac{4}{x+y}\ge\frac{4}{4}=1\)
Dấu " = " xảy ra <=> x=y=2
Hình như anh kudo shinichi ngược dấu một xíu thì phải ạ: \(8\ge\frac{\left(x+y\right)^2}{2}\Rightarrow\left(x+y\right)\le4\) chứ ạ?Dẫn đến
khúc sau ngược dấu.Nếu em sai thì xin thông ảm cho ạ. Lời giải của em đây:
\(A\ge\frac{4}{x+y}=\frac{16}{4x+4y}\ge\frac{16}{x^2+4+y^2+4}\) (BĐT Cô si hay AM-GM gì đó: \(x^2+4\ge2\sqrt{x^2.4}=2.2.x=4x;...\))
\(=\frac{16}{8+8}=1\).Dấu "=" xảy ra khi x = y = 2.
Vậy min A = 1 khi x =y = 2