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令\(x>y>0\)
则\(\frac{2x^3+2y^3}{x-y}=1\)
所以\(\frac{1-2x^2}{y^2}=\frac{\frac{2x^3+2y^3}{x-y}-2x^2}{y^2}=\frac{\frac{2x^3+2y^3-2x^3+2x^2y}{x-y}}{y^2}\)
\(=\frac{2y^3+2x^2y}{\left( x-y\right)y^2}=\frac{2\left( x^2+y^2\right)}{xy-y^2}=\frac{2\left[ 1+\left( \frac{x}{y}\right)^2\right]}{\frac{x}{y}-1}\)
令\(\frac{x}{y}=t>1\)
那么\(\frac{1-2x^2}{y^2}=\frac{2\left( 1+t^2\right)}{t-1}=2\frac{\left( t-1\right)^2+2t}{t-1}=2\frac{\left( t-1\right)^2+2\left( t-1\right)+2}{t-1}\)
\(=2\left[ \left( t-1\right)+\frac{2}{t-1}+2\right]=4+2\left( t-1+\frac{2}{t-1}\right)\ge4+4\sqrt{2}\)
所以 \(\left( \frac{1-2x^2}{y^2}\right)_{\min}=4+4\sqrt{2}\) |
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hbghlyj
发表于 2025-1-7 03:09
