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\(M=\frac{x^2}{xy}+\frac{y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\frac{x}{y}+\frac{1}{\frac{x}{y}}\)
\(x\ge2y\Rightarrow\frac{x}{y}\ge2\)
\(\Rightarrow M\ge2+\frac{1}{2}=\frac{5}{2}\)
GTNN của M là \(\frac{5}{2}\)khi \(a=2y\)
\(\frac{x}{y}>=2\)=>\(\frac{y}{x}=< \frac{1}{2}\)
\(M=\frac{x}{y}+\frac{y}{x}=\frac{x}{y}+\frac{4y}{x}-\frac{3y}{x}\)
ta có \(\frac{x}{y}+\frac{4y}{x}>=4\)(cô si)(1)
\(-\frac{3y}{x}>=-\frac{3}{2}\)(2)
cộng 1 với 2=>M>=5/2
xảy ra dâu = khi x/y=2
Ta có \(\frac{a}{a^2+2b+3}=\frac{a}{a^2+1+2\left(b+1\right)}\le\frac{a}{2a+2\left(b+1\right)}=\frac{a}{2\left(a+b+1\right)}\)
Chứng minh tương tự \(\hept{\begin{cases}\frac{b}{b^2+2c+3}\le\frac{b}{2\left(b+c+1\right)}\\\frac{c}{c^2+2a+3}\le\frac{c}{2\left(a+c+1\right)}\end{cases}}\)
Cộng 3 vế của 3 bđt lại ta được
\(VT\le\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)
Để bài toán được chứng minh thì ta cần \(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)
\(\Leftrightarrow1-\frac{a}{a+b+1}+1-\frac{b}{b+c+1}+1-\frac{c}{c+a+1}\ge2\)
\(\Leftrightarrow A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge2\)
Ta có \(A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\)
\(=\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\frac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\frac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\)
Áp dụng bđt quen thuộc \(\frac{m^2}{x}+\frac{n^2}{y}+\frac{p^2}{z}\ge\frac{\left(m+n+p\right)^2}{x+y+z}\)(quen thuộc) ta được
\(A\ge\frac{\left(a+b+c+3\right)^2}{\left(b+1\right)\left(a+b+1\right)+\left(c+1\right)\left(b+c+1\right)+\left(a+1\right)\left(c+a+1\right)}\)
\(=\frac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)
\(=\frac{2\left(a+b+c+3\right)^2}{2\left(a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\right)}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+6}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9}\)
\(=\frac{2\left(a+b+c+3\right)^2}{\left(a+b+c+3\right)^2}=2\)(DDpcm)
Dấu "=" xảy ra tại a= b = c =1
bn có thể ghi cho mk cái bđt đấy đc ko
#mã mã#
Bổ đề: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)
\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{\left[\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)\right]^2}{2}=\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
(Cứ thấy sao sao?x + y = 1 = > x = y = 1/2)
Với ĐK : x + y = 1 ... , chỉ có x = y = 1/2 (cái nài là STP mà có phải SD đâu??)
Chia làm 2TH
\(N>\frac{25}{2}\); TH2 : \(N=\frac{25}{2}\)
\(N=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{25}{2}\)
\(N=\left(\frac{1}{2}+\frac{1}{\frac{1}{2}}\right)^2+\left(\frac{1}{2}+\frac{1}{\frac{1}{2}}\right)^2\ge\frac{25}{2}\)
\(N=\left(\frac{1}{2}+1\div1\div2\right)^2+\left(\frac{1}{2}+1\div1\div2\right)^2\ge\frac{25}{2}\)
\(N=\left(\frac{1}{2}+1\div2\right)^2+\left(\frac{1}{2}+1\div2\right)^2\ge\frac{25}{2}\)
\(N=\left(\frac{1}{2}+\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{1}{2}\right)^2\ge\frac{25}{2}\)
\(N=\left(1\right)^2+\left(1\right)^2\ge\frac{25}{2}\)
\(N=2\ge\frac{25}{2}\)
----------------------------
\(N=\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\ge\frac{25}{2}\)
Tương tự như trên :\(N=\left(\frac{1}{2}+\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{1}{2}\right)^2\)
\(N=\left(\frac{1}{2}+\frac{1}{2}\right)\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)\ge\frac{25}{2}\)
Chẳng khác gì phía trên,mà 25 / 2 = 25 : 2 = 12 , 5 . Lại còn x , y là số dương .
[Trình mình thì chẳng CM được cái này(vì không CM được)]
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
easy!
Ta có:
\(\frac{1}{x^3\left(2y-x\right)}+x^2+y^2=\frac{1}{x^2\left(2xy-x^2\right)}+x^2+\left(y^2+x^2-x^2\right)\)
Áp dụng bất đẳng thức AM-GM cho hai số không âm,ta được:
\(x^2+y^2\ge2xy\)
\(\Rightarrow\frac{1}{x^3\left(2y-x\right)}+x^2+y^2\ge\frac{1}{x^2\left(2xy-x^2\right)}+x^2+\left(2xy-x^2\right)\)
Áp dụng bất đẳng thức AM-GM một lần nữa,ta được:
\(\frac{1}{x^3\left(2y-x\right)}+x^2+y^2\ge3\sqrt[3]{\frac{1}{x^2\left(2xy-x^2\right)}\cdot x^2\cdot\left(2xy-x^2\right)}=3\left(đpcm\right)\)
xong!
Áp dụng bđt quen thuộc \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a;b;c>0\right)\) được
\(\frac{1}{x}+\frac{2}{y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\ge\frac{9}{x+2y}=\frac{9}{3}=3\)
Dấu "=" tại x = y = 1