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2) Ta có:
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\)
Áp dụng BĐT Schwarz:
\(\frac{1}{2xy}+\frac{1}{x^2+y^2}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\)
Mà x+y=1 nên suy ra:
\(\frac{1}{2xy}+\frac{1}{x^2+y^2}\ge4\)
\(\Rightarrow2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\ge8\)
=>đpcm.
Dấu ''='' xảy ra khi x=y=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left(x-1;y-1\right)=\left(a;b\right)\Rightarrow\left(x;y\right)=\left(a+1;b+1\right)\)
\(VT=\dfrac{\left(a+1\right)^3+\left(b+1\right)^3-\left(a+1\right)^2-\left(b+1\right)^2}{ab}=\dfrac{a^3+a+b^3+b+2\left(a^2+b^2\right)}{ab}\)
\(VT\ge\dfrac{2a^2+2b^2+2\left(a^2+b^2\right)}{ab}=\dfrac{4\left(a^2+b^2\right)}{ab}\ge\dfrac{8ab}{ab}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=\frac{2}{2xy}+\frac{2}{x^2+y^2}\ge\frac{2.4}{2xy+x^2+y^2}=\frac{8}{\left(x+y\right)^2}=8\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
\(\frac{x^2}{y-1}+\frac{y^2}{x-1}=\frac{x^2}{y-1}+4.\left(y-1\right)+\frac{y^2}{x-1}+4.\left(x-1\right)-4x-4y+8\)
Áp dụng BĐT AM-GM ta có:
\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge2.\sqrt{\frac{x^2}{y-1}.4\left(y-1\right)}+2.\sqrt{\frac{y^2}{x-1}.4.\left(x-1\right)}-4x-4y+8\)
\(=2.\sqrt{4x^2}+2.\sqrt{4.y^2}-4x-4y+8\)
\(=4x+4y-4x-4y+8\)
\(=8\)
đpcm