The solution of X inequality: ax + X / 1 > 1, is 1 / X

The solution of X inequality: ax + X / 1 > 1, is 1 / X

ax+1/x>1
(ax^2+1)/x>1
When x > 0, ax ^ 2 + 1 > x is ax ^ 2-x + 1 > 0
a> The discriminant of equation AX ^ 2-x + 1 = 0 is Δ = 1-4a
If Δ ≥ 0, i.e. a ≤ 1 / 4, the solution of the inequality is
x[1+√(1-4a)]/(2a)
If Δ ≤ 0, i.e. a ≥ 1 / 4, the solution of the inequality is r
To sum up, 0

x. Y ∈ (0, + ∞), x + 2Y + xy = 30, find the value range of X + y
The answer is x + y ∈ [8 √ 2-3,30],

x+2y+xy=30,
Then y = (30-x) / (x + 2),
Because y & gt; 0, & nbsp; (30-x) / (x + 2) & nbsp; & gt; & nbsp; 0,
So 0 & lt; X & lt; 30
Let x + 2 = t, then x = t-2,2 & lt; T & lt; 32
x+y= x+(30-x)/(x+2)
=t-2+(32-t)/t
=t-2+32/t-1 
= t+32/t-3…… Using basic inequality
≥2√(t•32/t)-3=8√2-3.
The function T + 32 / T decreases on [0,4 √ 2] and increases on [& nbsp; 4 √ 2, + ∞),
Because of 2 & lt; T & lt; 32, when t = 32, t + 32 / T is the largest, t + 32 / T-3 & lt; 30
∴x+y∈[8√2-3,30).