202601163

$\mathrm{若}\mathrm{实}\mathrm{数}x,y\mathrm{满}\mathrm{足}{x}^2\ge 3+y^2,\mathrm{求}\frac{3}{x^2}+\frac{4y}{x}\mathrm{的}\mathrm{最}\mathrm{大}\mathrm{值}$

$\mathrm{由}\mathrm{已}\mathrm{知}\mathrm{可}\mathrm{得}3\le x^2-y^2$
${\displaystyle \frac{3}{x^2}}+{\displaystyle \frac{4y}{x}}\le {\displaystyle \frac{x^2-y^2}{x^2}}+{\displaystyle \frac{4y}{x}}=1-{\displaystyle \frac{y^2}{x^2}}+{\displaystyle \frac{4y}{x}}$
$x\ne 0,\mathrm{令}t={\displaystyle \frac{y}{x}}$
$1-{\displaystyle \frac{y^2}{x^2}}+{\displaystyle \frac{4y}{x}}=t^2+4t-1=-(t-2{)}^2+3\le 3$