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Ta có:
\(15\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=10\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+2014\)
\(\le10\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2014\)
=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le\frac{2014}{5}\)
\(P=\frac{1}{\sqrt{5x^2+2xy+2yz}}+\frac{1}{\sqrt{5y^2+2yz+2zx}}+\frac{1}{\sqrt{5z^2+2zx+2xy}}\)
=> \(P\sqrt{\frac{2014}{135}}=\frac{1}{\sqrt{5x^2+2xy+2yz}.\sqrt{\frac{135}{2014}}}\)
\(+\frac{1}{\sqrt{5y^2+2yz+2zx}\sqrt{\frac{135}{2014}}}+\frac{1}{\sqrt{\frac{135}{2014}}\sqrt{5z^2+2zx+2xy}}\)
\(\le\frac{1}{2}\left(\frac{1}{5x^2+2xy+2yz}+\frac{2014}{135}+\frac{1}{5y^2+2yz+2zx}+\frac{2024}{135}+\frac{1}{5z^2+2yz+2zx}+\frac{2014}{135}\right)\)
\(\le\frac{1}{2}\left[\frac{1}{81}\left(\frac{5}{x^2}+\frac{2}{xy}+\frac{2}{yz}\right)+\frac{1}{81}\left(\frac{5}{y^2}+\frac{2}{yz}+\frac{2}{zx}\right)+\frac{1}{81}\left(\frac{5}{z^2}+\frac{2}{zx}+\frac{2}{xy}\right)+\frac{2014}{45}\right]\)
\(=\frac{5}{162}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2}{81}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{1007}{45}\)
\(\le\frac{5}{162}.\frac{2014}{5}+\frac{2}{81}.\frac{2014}{5}+\frac{1007}{45}=\frac{2014}{45}\)
=> \(P\le\frac{2014}{45}:\sqrt{\frac{2014}{135}}=3\sqrt{\frac{2014}{135}}\)
Dấu "=" xảy ra <=> x = y = z = \(\sqrt{\frac{15}{2014}}\)
Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)
Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)
=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)
từ giả thiết ta suy ra \(\sqrt[3]{x^2y^2z^2}\ge3\)
lại có x2 + 2yz = x2 + yz + yz \(\ge\)3\(\sqrt[3]{x^2y^2z^2}\)\(\ge\)9
nên \(\frac{1}{x^2+2yz}\le\frac{1}{9}\)
tương tự với 2 số còn lại nên ta được P \(\le\frac{1}{3}\)
dấu "=" xảy ra khi x = y = z = \(\sqrt{3}\)
Lời giải:
Xét mẫu thức:
$2xy^2+2yz^2+2zx^2+3xyz=(xy^2+yz^2+zx^2)+(xy^2+xyz)+(yz^2+xyz)+(xz^2+xyz)$
$=xy^2+yz^2+zx^2+xy(y+z)+yz(z+x)+xz(x+y)$
$=xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)$
$=(x-y)(y-z)(z-x)$
$\Rightarrow (2xy^2+2yz^2+2zx^2)^2=(x-y)^2(y-z)^2(z-x)^2$
Xét tử thức:
$(xy+2z^2)(yz+2x^2)(xz+2y^2)$
$=[xy+z^2-z(x+y)][yz+x^2-x(z+y)][xz+y^2-y(x+z)]$
$=(z-x)(z-y)(x-y)(x-z)(y-x)(y-z)=-(x-y)^2(y-z)^2(z-x)^2$
Do đó: $A=-1$
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
