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Câu 1:
\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)
\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)
Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)
\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)
Dấu = xảy ra khi x=y=1/2
Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)
CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)
Dấu = xảy ra khi x=y=z=1
Ta có: \(6x^2+8xy+11y^2=2\left(x-y\right)^2+\left(2x+3y\right)^2\ge\left(2x+3y\right)^2\)
Tương tự: \(6y^2+8yz+11z^2\ge\left(2y+3z\right)^2\)
\(6z^2+8zx+11x^2\ge\left(2z+3x\right)^2\)
=> \(P\le\frac{x^2+3xy+y^2}{2x+3y}+\frac{y^2+3yz+z^2}{2y+3z}+\frac{z^2+3zx+x^2}{2z+3x}\)
=> \(4P\le\frac{4x^2+12xy+4y^2}{2x+3y}+\frac{4y^2+12yz+4z^2}{2y+3z}+\frac{4z^2+12zx+4x^2}{2z+3x}\)
\(=\frac{\left(2x+3y\right)^2-5y^2}{2x+3y}+\frac{\left(2y+3z\right)^2-5z^2}{2y+3z}+\frac{\left(2z+3x\right)^2-5x^2}{2z+3x}\)
\(=5\left(x+y+z\right)-5\left(\frac{y^2}{2x+3y}+\frac{z^2}{2y+3z}+\frac{x^2}{2z+3x}\right)\)
\(\le5\left(x+y+z\right)-5.\frac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=4\left(x+y+z\right)\)
Lại có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)với mọi x; y; z
=> \(4P\le4.\sqrt{9}=12\)
=> \(P\le3\)
Dấu "=" xảy ra <=> x = y = z = 1
Vậy max P = 3 đạt tại x = y = z = 1.
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 \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
\(\text{Áp dụng BĐT:}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)
\(\frac{1}{3x+3y+2z}=\frac{1}{\left(x+y\right)+\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(\text{tương tự với các BĐT còn lại }\)
\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+3z+2y}+\frac{1}{3y+3z+2x}\le\frac{1}{16}.\left(\frac{4}{x+z}+\frac{4}{x+y}+\frac{4}{y+z}\right)=\frac{1}{16}.24=\frac{3}{2}\left(đpcm\right)\)
Đặt \(\hept{\begin{cases}x=a\\2y=b\\3z=c\end{cases}}\left(a;b;c>0\right)\Rightarrow a+b+c=2\)
Khi đó \(S=\Sigma\sqrt{\frac{\frac{ab}{2}}{\frac{ab}{2}+c}}=\Sigma\sqrt{\frac{ab}{ab+2c}}=\Sigma\sqrt{\frac{ab}{ab+\left(a+b+c\right)c}}\)
\(=\Sigma\sqrt{\frac{ab}{ab+bc+ca+c^2}}=\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
Áp dụng bđt Cô-si có
\(S\le\frac{\Sigma\left(\frac{a}{a+c}+\frac{b}{b+c}\right)}{2}=\frac{3}{2}\)
a) \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2\left(y^2+\frac{1}{x^2}\right)\)
\(+\frac{1}{y^2}\left(y^2+\frac{1}{x^2}\right)=x^2y^2+2+\frac{1}{x^2y^2}\)
\(=2+\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}\)
Áp dụng BĐT Cauchy - Schwar cho 2 số không âm, ta được:
\(x^2y^2+\frac{1}{256x^2y^2}\ge2\sqrt{\frac{x^2y^2}{256x^2y^2}}=\frac{1}{8}\)
C/m được BĐT phụ: \(1=\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow16x^2y^2\le1\Leftrightarrow256x^2y^2\le16\Leftrightarrow\frac{255}{256x^2y^2}\ge\frac{255}{16}\)
\(\Rightarrow M\ge2+\frac{1}{8}+\frac{255}{16}=\frac{289}{16}\)
(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x^2y^2=\frac{1}{256x^2y^2}\\x-y=0\end{cases}}\Leftrightarrow x=y=\frac{1}{2}\))
\(\frac{16}{3x+3y+2z}=\frac{16}{\left(x+y\right)+\left(y+z\right)+\left(z+x\right)+\left(x+y\right)1}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\)
Tương tự \(\frac{16}{3x+2y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+z}\)
\(\frac{16}{2x+3y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{y+z}\)
Cộng vế theo vế ta có:
\(16\left(\frac{1}{3x+2y+3z}+\frac{1}{3x+3y+2z}+\frac{1}{2x+3y+3z}\right)\le4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=24\)
\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\left(đpcm\right)\)
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Ta có:
\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\ge\frac{9}{2\left(x+y+z\right)}\)\(\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có: \(\frac{1}{2x+3y+3z}=\frac{\left(\frac{3}{4}+\frac{1}{4}\right)^2}{2\left(x+y+z\right)+y+z}\le\frac{9}{32\left(x+y+z\right)}+\frac{1}{16\left(y+z\right)}\)
Do đó:
\(\frac{1}{2x+3y+3z}+\frac{1}{2y+3x+3z}+\frac{1}{2z+3x+3y}\)
\(\le\frac{9}{32\left(x+y+z\right)}\cdot3+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(\le\frac{9}{32\cdot\frac{3}{4}}+\frac{1}{16}\cdot6=\frac{3}{2}\)(Đpcm)
ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)
Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)
\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla