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Ta có \(\left(\frac{x^3}{y^2+z}+\frac{y^3}{z^2+x}+\frac{z^3}{x^2+y}\right)\left[x\left(y^2+x\right)+y\left(z^2+x\right)+z\left(x^2+y\right)\right]\ge\left(x^2+y^2+z^2\right)^2\left(1\right)\)
Ta chứng minh \(\left(x^2+y^2+z^2\right)^2\ge\frac{4}{5}\left[x\left(y^2+z\right)+y\left(z^2+x\right)+z\left(x^2+y\right)\right]\)
\(\Leftrightarrow5\left(x^2+y^2+z^2\right)^2\ge4\left[x\left(y^2+z\right)+y\left(z^2+x\right)+z\left(x^2+y\right)\right]\left(2\right)\)
Thật vậy \(\hept{\begin{matrix}3\left(\Sigma x^2\right)^2\ge\left(\Sigma x^2\right)\cdot\Sigma x^2=4\Sigma zx\left(3\right)\\2\left(\Sigma x^2\right)^2\ge4\Sigma xy^2\left(4\right)\end{matrix}\Leftrightarrow2\left(\Sigma x^2\right)^2\ge\Sigma xy^2\left(x+y+z\right)}\)(*)
Từ các Bất Đẳng Thức \(\hept{\begin{cases}\frac{x^4-2x^3z+z^2x^2}{2}\ge0\\\frac{x^4+y^4+2x^4}{4}\ge xyz^2\end{cases}}\)=> (*) đúng
Như vậy (3),(4) đúng => (2) đúng
Từ đó suy ra \(T\ge\frac{4}{5}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)
Theo giả thiết ta có : \(x+yz=yz-z-1=\left(z-1\right)\left(y+1\right)=\left(x+y\right)\left(y+1\right)\)
Tương tự : \(y+zx=\left(x+y\right)\left(x+1\right)\)
Và \(z+xy=\left(x+1\right)\left(y+1\right)\)
Nên \(P=\frac{x}{\left(x+y\right)\left(y+1\right)}+\frac{y}{\left(x+y\right)\left(x+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
\(=\frac{x^2+y^2+x+y}{\left(x+y\right)\left(x+1\right)\left(y+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
Ta có \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\left(x+1\right)\left(y+1\right)\le\frac{\left(x+y+2\right)^2}{4}\)
nên \(P\ge\frac{2\left(x+y\right)^2+4\left(x+y\right)}{\left(x+y+2\right)^2\left(x+y\right)}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}=\frac{2\left(x+y\right)+4}{\left(x+y+2\right)^2}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}\)
\(=\frac{2}{z+1}+\frac{4\left(z^2+2\right)}{\left(z+1\right)^2}=f\left(z\right);z>1\)
Lập bảng biến thiên ta được \(f\left(z\right)\ge\frac{13}{4}\) hay min \(P=\frac{13}{4}\) khi \(\begin{cases}z=3\\x=y=1\end{cases}\)
\(H=\sum\frac{y}{x^2+1+2y+2}\le\sum\frac{y}{2x+2y+2}=\frac{1}{2}\sum\frac{y}{x+y+1}\)
Ta sẽ chứng minh \(H\le\frac{1}{2}\) hay \(\frac{y}{x+y+1}+\frac{z}{y+z+1}+\frac{x}{z+x+1}\le1\)
\(\Leftrightarrow\frac{x+1}{x+y+1}+\frac{y+1}{y+z+1}+\frac{z+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x+y+1\right)}+\frac{\left(y+1\right)^2}{\left(y+1\right)\left(y+z+1\right)}+\frac{\left(z+1\right)^2}{\left(z+1\right)\left(z+x+1\right)}\)
\(VT\ge\frac{\left(x+y+z+3\right)^2}{\left(x+1\right)\left(x+y+1\right)+\left(y+1\right)\left(y+z+1\right)+\left(z+1\right)\left(z+x+1\right)}\)
\(VT\ge\frac{\left(x+y+z+3\right)^2}{x^2+y^2+z^2+xy+yz+zx+3x+3y+3z+3}=\frac{\left(x+y+z+3\right)^2}{\frac{1}{2}\left(x^2+y^2+z^2\right)+xy+yz+zx+3x+3y+3z+3+\frac{1}{2}\left(x^2+y^2+z^2\right)}\)
\(VT\ge\frac{\left(x+y+z+3\right)^2}{\frac{1}{2}\left(x+y+z\right)^2+3\left(x+y+z\right)+3+\frac{3}{2}}=\frac{\left(x+y+z+3\right)^2}{\frac{1}{2}\left(x+y+z\right)^2+3\left(x+y+z\right)+\frac{9}{2}}\)
\(VT\ge\frac{\left(x+y+z+3\right)^2}{\frac{1}{2}\left(x+y+z+3\right)^2}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng bất đẳng thức Minkowski ta có:
\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
\(=\sqrt{\left[\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}\right]+\frac{80}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{2\sqrt{\left(x+y+z\right)^2\cdot\frac{1}{\left(x+y+z\right)^2}}+\frac{80}{1}}=\sqrt{82}\)
Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{3}\)
Áp dụng bất đẳng thức Minkowski ta có:
√x2+1x2 +√y2+1y2 +√z2+1z2 ≥√(x+y+z)2+(1x +1y +1z )2
≥√(x+y+z)2+(9x+y+z )2=√(x+y+z)2+81(x+y+z)2
=√[(x+y+z)2+1(x+y+z)2 ]+80(x+y+z)2
≥√2√(x+y+z)2·1(x+y+z)2 +801 =√82
Dấu "=" xảy ra khi: x=y=z=13
Áp dụng bất đẳng thức Cauchy
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+zx}\)
\(M\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+xz\right)}+\frac{7}{xy+yz+zx}\)
Áp dụng BĐT Cauchy - Schwarz :
\(\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}\ge\frac{\left(1+2\right)^2}{\left(x+y+z\right)^2}=9\)
và \(\frac{7}{xy+yz+xz}\ge\frac{7}{\frac{1}{3}\left(x+y+z\right)^2}=21\)
\(\Rightarrow M\ge9+21=30\)
Dấu " = " xảy ra khi \(x=y=z=\frac{1}{3}\)
Áp dụng BĐT Cauchy schwarz ta có:
\(M=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
\(\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}\)
\(=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}+\frac{7}{2\left(xy+yz+zx\right)}\)
\(\ge\frac{9}{\left(x+y+z\right)^2}+\frac{7}{\frac{2\left(x+y+z\right)^2}{3}}=30\)
Đẳng thức xảy ra tại x=y=z=1/3
\(VT=x^2+y^2+z^2+\frac{49}{x+2y+3z}\)
\(=\frac{x^2}{1}+\frac{\left(2y\right)^2}{4}+\frac{\left(3y\right)^2}{9}+\frac{49}{\left(x+2y+3z\right)}\)
\(\ge\frac{\left(x+2y+3z\right)^2}{14}+\frac{49}{2\left(x+2y+3z\right)}+\frac{49}{2\left(x+2y+3z\right)}\)
\(\ge3\sqrt[3]{\frac{\left(x+2y+3z\right)^2}{14}.\frac{49}{2\left(x+2y+3z\right)}.\frac{49}{2\left(x+2y+3z\right)}}=\frac{21}{2}\)
Đẳng thức xảy ra khi \(x=\frac{1}{2};y=1;z=\frac{3}{2}\)
