X=2t-1/t , y=t+4/t (t is not 0) deduce that dy/dx<1/2

dy/dx = (dy/dt) / (dx/dt)
.........= (1 - 4/t^2) / (2 + 1/t^2)
.........= (t^2 - 4) / (2t^2 + 1)
.........= (t^2 + 1/2 - 9/2) / (2t^2 + 1)
.........= [(1/2)(2t^2 + 1) - 9/2] / (2t^2 + 1)
.........= 1/2 - (9/2)/(2t^2 + 1)
.........< 1/2, since (9/2)/(2t^2 + 1) > 0 for all t.

I hope this helps!

dx/dt = 2+1/t^2 = (2t^2+1)/t^2
dy/dt = 1-4/t^2 = (t^2-4)/t^2
dy/dx = (t^2-4)/(2t^2+1)
dy/dx = (1-4/t^2)/(2+1/t^2)
t^2 is always positive
So dy/dx < 1/2