Determine the largest and smallest fractions $F = \frac{y-x}{x+4y}$ if the real numbers $x$ and $y$ satisfy the equation $x^2y^2 + xy + 1 = 3y^2$.
Problem
Source: Czech-Polish-Slovak Junior Match 2014, Team p6 CPSJ
Tags: inequalities, algebra, minimum, maximum, min, max
23.03.2020 15:37
Hello, it is $$\frac{y-x}{x+4y}\le 4$$the equal sign holds if $$x=\sqrt{3},y=-\frac{1}{\sqrt{3}}$$
23.03.2020 15:41
and we get $$\frac{y-x}{x+4y}\geq 0$$the equal sign holds if $$x=y=-1$$
23.03.2020 16:13
Let $x$ and $y$ be real numbers satisfy $x^2y^2 + xy + 1 = 3y^2.$ Prove that $$0\leq \frac{y-x}{x+4y}\leq 4.$$Let $x$ and $y$ be real numbers satisfy $7x^2 + 3xy + 3y^2 = 1.$ Prove that $$\frac{x^2 +y^2}{y}\geq \frac{1}{2}$$(BMO 1988)
11.11.2021 12:44
For P1 FOR MAXIMUM $y-x<=4x+16y$ $-15y<=5x$ $-3y<=x$ FOR MINIMUM $x<=y$
11.11.2021 14:52
StarLex1 wrote: For P1 FOR MAXIMUM $y-x<=4x+16y$ $-15y<=5x$ $-3y<=x$ FOR MINIMUM $x<=y$ No.