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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
\(P=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{xy+xz+yz}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{1}{x}+\dfrac{y^2}{2}+\dfrac{1}{y}+\dfrac{z^2}{2}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\left(\dfrac{x^2}{2}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)+\left(\dfrac{y^2}{2}+\dfrac{1}{2y}+\dfrac{1}{2y}\right)+\left(\dfrac{z^2}{2}+\dfrac{1}{2z}+\dfrac{1}{2z}\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{x^2}{2}.\dfrac{1}{2x}.\dfrac{1}{2x}}+3\sqrt[3]{\dfrac{y^2}{2}.\dfrac{1}{2y}.\dfrac{1}{2y}}+3\sqrt[3]{\dfrac{z^2}{2}.\dfrac{1}{2z}.\dfrac{1}{2z}}=\dfrac{9}{2}\)
\(\Rightarrow P_{min}=\dfrac{9}{2}\) khi \(x=y=z=1\)
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 \(\begin{cases}z=3\\x=y=1\end{cases}\)
Ta có: \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge9xyz\)
\(VT=\dfrac{x}{1+yz}+\dfrac{y}{1+xz}+\dfrac{z}{1+xy}\)
\(=\dfrac{x^2}{x+xyz}+\dfrac{y^2}{y+xyz}+\dfrac{z^2}{z+xyz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3xyz}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\dfrac{\left(x+y+z\right)\left(xy+yz+xz\right)}{3}}\)
\(=\dfrac{3\left(x+y+z\right)}{4}\). Cần chứng minh:
\(\dfrac{3\left(x+y+z\right)}{4}\ge\dfrac{3\sqrt{3}}{4}\Leftrightarrow x+y+z\ge\sqrt{3}\)
BĐT cuối đúng vì \(x+y+z\ge\sqrt{3\left(xy+yz+xz\right)}=\sqrt{3}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Ps: nospoiler
Lời giải:
Từ \(x+y-z=-1\Rightarrow z-x-y=1\)
Ta có các biến đổi sau:
\(x+yz=x(z-x-y)+yz=x(z-x)+y(z-x)=(x+y)(z-x)\)
\(=(x+y)(y+1)\)
\(y+zx=y(z-x-y)+zx=y(z-y)+x(z-y)=(y+x)(z-y)\)
\(=(y+x)(x+1)\)
\(z+xy=z(z-x-y)+xy=(z-x)(z-y)=(x+1)(y+1)\)
Khi đó:\(P=\frac{x^3y^3}{(x+y)^2(x+1)^3(y+1)^3}(*)\)
Áp dụng BĐT Cauchy:
\((x+y)^2\geq 4xy\)
\(x+1=\frac{x}{2}+\frac{x}{2}+1\geq 3\sqrt[3]{\frac{x^2}{4}}\Rightarrow (x+1)^3\geq \frac{27x^2}{4}\)
\(y+1\geq 3\sqrt[3]{\frac{y^2}{4}}\Rightarrow (y+1)^3\geq \frac{27y^2}{4}\) (tương tự ở trên)
\(\Rightarrow (x+y)^2(x+1)^3(y+1)^3\geq \frac{729}{4}x^3y^3(**)\)
Từ \((*); (**)\Rightarrow P\leq \frac{x^3y^3}{\frac{729}{4}x^3y^3}=\frac{4}{279}\Rightarrow P_{\max}=\frac{4}{729}\)
Đẳng thức xảy ra khi \(x=y=2; z=5\)
\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)
\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)
Do đó:
\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)
\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)
Đẳng thức xảy ra khi $x=y=z=1.$
Ta xét BĐT phụ: \(1+x^3+y^3\ge xy\left(x+y+z\right)\)
\(x^3+y^3\ge xy\left(x+y\right)+xyz-1\)
\(x^3+y^3-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x-y\right)^2\ge0\)( Luôn đúng, vậy BĐT phụ đúng)
\(\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}=\sqrt{x+y+z}.\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3\sqrt[3]{xyz}}.\left(3\sqrt[3]{\dfrac{1}{\sqrt{x^2y^2z^2}}}\right)=3\sqrt{3}\)
GTNN của P là \(3\sqrt{3}\Leftrightarrow x=y=z=1\)
Ta có:
\(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}=\dfrac{1}{2}\left(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{x}{yz}+\dfrac{z}{xy}+\dfrac{y}{zx}+\dfrac{z}{xy}\right)\ge\dfrac{1}{2}\left(\dfrac{2}{z}+\dfrac{2}{y}+\dfrac{2}{x}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+y^2+z^2\right)+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+\dfrac{1}{x}+\dfrac{1}{x}\right)+\dfrac{1}{2}\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)+\dfrac{1}{2}\left(z^2+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
\(\Rightarrow P\ge\dfrac{3}{2}\sqrt[3]{\dfrac{x^2}{x^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{y^2}{y^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{z^2}{z^2}}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT cauchy:
\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}\)
\(M\ge\dfrac{1}{x^2+y^2+z^2}+\dfrac{9}{xy+yz+xz}=\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+xz\right)}+\dfrac{7}{xy+yz+zx}\)Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+zx\right)}\ge\dfrac{\left(1+2\right)^2}{\left(x+y+z\right)^2}=9\)
và \(\dfrac{7}{xy+yz+xz}\ge\dfrac{7}{\dfrac{1}{3}\left(x+y+z\right)^2}=21\)
\(\Rightarrow M\ge9+21=30\)
dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)
cô si cho đễ hiểu đi bn , cần gì phải cauchy s,. làm gì cho mệt