Question

Виконайте завдання 10. в)

Answer 1

$3(\frac{x^2}{y^2} +\frac{y^2}{x^2} )-8(\frac{x}{y} +\frac{y}{x} )+10\geq 0\\ \\ 3((\frac{x}{y} +\frac{y}{x} )^2-2)-8(\frac{x}{y} +\frac{y}{x} )+10\geq 0\\ \\ 3(\frac{x}{y} +\frac{y}{x} )^2-8(\frac{x}{y} +\frac{y}{x} )+4\geq 0\\ \\D=64-4*3*4=16=4^2 \\ \\(\frac{x}{y} +\frac{y}{x} )=(8+4)/6=2\\ \\(\frac{x}{y} +\frac{y}{x} )=(8-4)/6=2/3 \\ \\  3(\frac{x}{y} +\frac{y}{x}-2 )(\frac{x}{y} +\frac{y}{x}- \frac{2}{3} )\geq 0\\ \\ 3*\frac{x^2+y^2-2xy}{xy} *\frac{3x^2+3y^2-2xy}{3xy} \geq 0$

$\frac{(x-y)^2(3x^2-2xy+3y^2)}{x^2y^2} \geq 0\\ \\ ( x-y)^2\geq 0;x^2y^2> 0;(x\neq 0;y\neq 0)\\ \\ 3x^2-2xy+3y^2>0;t.k.\ D=4y^2-4*3*3y^2=-32y^2<0$

неравенство верное, ч.т.д