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\(A=\dfrac{x^3+y^3+4}{xy+1}\ge\dfrac{x^3+y^3+4}{\dfrac{x^2+y^2}{2}+1}=\dfrac{x^3+y^3+4}{2}=\dfrac{\dfrac{1}{2}\left(x^3+x^3+1\right)+\dfrac{1}{2}\left(y^3+y^3+1\right)+3}{2}\)
\(\ge\dfrac{\dfrac{3}{2}\left(x^2+y^2\right)+3}{2}=3\)
\(A_{min}=3\) khi \(x=y=1\)
Do \(x^2+y^2=2\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{2}\\y\le\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3\le\sqrt{2}x^2\\y^3\le\sqrt{2}y^2\end{matrix}\right.\)
\(\Rightarrow A\le\dfrac{\sqrt{2}\left(x^2+y^2\right)+4}{xy+1}=\dfrac{4+2\sqrt{2}}{xy+1}\le\dfrac{4+2\sqrt{2}}{1}=4+2\sqrt{2}\)
\(A_{max}=4+2\sqrt{2}\) khi \(\left(x;y\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)
\(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)
Dấu "=" <=> x= y = 1/2
\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)
\(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)
Dấu "=" <=> x = 3y
![](https://rs.olm.vn/images/avt/0.png?1311)
\(BĐT\Leftrightarrow\frac{2-2xy}{2+x^2+y^2}+\frac{2x^2-2y}{1+2x^2+y^2}+\frac{2y^2-2x}{1+x^2+2y^2}\ge0\)
\(\Leftrightarrow1-\frac{2-2xy}{2+x^2+y^2}+1-\frac{2x^2-2y}{1+2x^2+y^2}+1-\frac{2y^2-2x}{1+x^2+2y^2}\le3\)
\(\Leftrightarrow\frac{\left(x+y\right)^2}{2+x^2+y^2}+\frac{\left(y+1\right)^2}{1+2x^2+y^2}+\frac{\left(x+1\right)^2}{1+x^2+2y^2}\le3\)(*)
Theo bất đẳng thức Bunyakovsky dạng phân thức: \(\frac{\left(x+y\right)^2}{2+x^2+y^2}\le\frac{x^2}{1+x^2}+\frac{y^2}{1+y^2}\)(1); \(\frac{\left(y+1\right)^2}{1+2x^2+y^2}\le\frac{y^2}{x^2+y^2}+\frac{1}{x^2+1}\)(2); \(\frac{\left(x+1\right)^2}{1+x^2+2y^2}\le\frac{x^2}{x^2+y^2}+\frac{1}{y^2+1}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{\left(x+y\right)^2}{2+x^2+y^2}+\frac{\left(y+1\right)^2}{1+2x^2+y^2}+\frac{\left(x+1\right)^2}{1+x^2+2y^2}\le\)\(\left(\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}\right)+\left(\frac{1}{y^2+1}+\frac{y^2}{y^2+1}\right)+\left(\frac{1}{x^2+1}+\frac{x^2}{x^2+1}\right)=3\)
Như vậy (*) đúng
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi x = y = 1
\(\frac{1-xy}{2+x^2+y^2}+\frac{x^2-y^2}{1+2x^2+y^2}+\frac{y^2-x}{1+x^2+2y^2}\ge0\)
\(\Leftrightarrow\frac{1-xy+3x^2-2y^2-2y^2+x}{\left(1+x^2+y^2\right)}\ge0\)
\(\Leftrightarrow\frac{2\left(1+x^2+y^2\right)+x^2}{1+x^2+y^2}\ge0\)
Vì x2 và y2 >0
\(\Rightarrow2+\frac{x^2}{1+x^2+y^2}\ge0\)(luôn đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
+) Đặt \(Q=\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)
\(Q=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}+\frac{xy-1}{xy-1}\)
\(Q=\frac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}+xy-1}{xy-1}\)
\(Q=\frac{-2-2\sqrt{x}}{xy-1}\)
\(Q=\frac{-2\left(\sqrt{x}+1\right)}{xy-1}\)
+) Đặt \(K=1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)
\(K=\frac{xy-1}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}\)
\(K=\frac{xy-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{xy-1}\)
\(K=\frac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)
\(K=\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)
Ta có : \(P=Q:K\)
\(\Leftrightarrow P=\frac{-2\left(\sqrt{x}+1\right)}{xy-1}:\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)
\(\Leftrightarrow P=\frac{-2\left(\sqrt{x}+1\right)\left(xy-1\right)}{-2\sqrt{xy}\left(\sqrt{x}+1\right)\left(xy-1\right)}\)
\(\Leftrightarrow P=\frac{1}{\sqrt{xy}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chỉ có biến đổi tương đương:
\(\frac{x^2+y^2+2}{\left(1+x^2\right)\left(1+y^2\right)}\le\frac{2}{1+xy}\Leftrightarrow\left(1+xy\right)\left(x^2+y^2+2\right)\le2\left(1+x^2\right)\left(1+y^2\right)\)
\(\Leftrightarrow x^2+y^2+2+x^3y+xy^3+2xy\le2+2x^2+2y^2+2x^2y^2\)
\(\Leftrightarrow xy\left(x^2+y^2-2xy\right)-\left(x^2-2xy+y^2\right)\le0\)
\(\Leftrightarrow\left(xy-1\right)\left(x-y\right)^2\le0\) (luôn đúng với mọi \(xy\le1\))
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}x=y\\xy=1\end{matrix}\right.\)
b/ Tính chất của z ở câu b là gì bạn? z bất kì là ko được đâu, hơn nữa mẫu số của vế phải thấy hơi kì quặc