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Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)
\(\Leftrightarrow P=\dfrac{2sin^2a+12sina.cosa}{1+2sina.cosa+2cos^2a}=\dfrac{1-cos2a+6sin2a}{2+sin2a+cos2a}\)
\(\Leftrightarrow P\left(2+sin2a+cos2a\right)=1-cos2a+6sin2a\)
\(\Leftrightarrow\left(P-6\right)sin2a+\left(P+1\right)cos2a=1-2P\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(P-6\right)^2+\left(P+1\right)^2\ge\left(1-2P\right)^2\)
\(\Leftrightarrow P^2+3P-18\le0\Rightarrow-6\le P\le3\)
Vậy \(\left\{{}\begin{matrix}P_{max}=3\\P_{min}=-6\end{matrix}\right.\)
\(\left(x+y\right)^3+4xy-2\ge0\) (1)
Ta có: \(\left(x-y\right)^2\ge0\) \(\forall x;y\)
\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow x^2+y^2+2xy-4xy\ge0\)
\(\Leftrightarrow\left(x+y\right)^2-4xy\ge0\) (2)
Cộng vế với vế của (1) và (2) ta được:
\(\left(x+y\right)^3+\left(x+y\right)^2-2\ge0\)
\(\Leftrightarrow S^3+S^2-2\ge0\Leftrightarrow\left(S-1\right)\left(S^2+2S+2\right)\ge0\)
Mặt khác \(S^2+2S+2=\left(S+1\right)^2+1>0\) \(\forall S\)
\(\Rightarrow S-1\ge0\Rightarrow S\ge1\)
\(\Rightarrow S_{min}=1\) khi \(x=y=\frac{1}{2}\)
\(P=3\left(x^2+y^2\right)^2-3x^2y^2-2\left(x^2+y^2\right)+1\)
\(\ge3\left(x^2+y^2\right)^2-\dfrac{3}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)
Đặt \(x^2+y^2=a\) thì \(a\ge2\).Xét hàm \(f\left(a\right)=\dfrac{9}{4}a^2-2a+1\)
Dế thấy \(f_{(a)}\) đồng biến trên [2,+\(\infty\)] nên \(f_{Min}\)=\(f_{(2)}\)=6
Dấu = xảy ra khi x=y=1