Solving inequalities in arithmetic

Question

If

$$S_S \ge \frac{120x}{\frac{60}{x}+1}$$

and

$2x\le 45$

then what can be said about the range of $S_S$?

Answer 1

For positive $x$, the numerator is a growing function and the denominator a decreasing one, so that the ratio is growing.

Hence,

$$\frac{120\cdot0}{\frac{60}{0}+1}\le S_S\le\frac{120\frac{45}2}{60\frac2{45}+1}.$$

(Though the lower bound is meaningless as written, the limit is $0$.)

Answer 2

See this.

$$S_s(a)>S_s(b) \space \text{for} \space a,b>0, a>b.$$

Hence the max on the positive is $S_s(22.5)=\frac{8100}{11}$

We see $x=0$ cannot exist due to $\frac{60}{x}$ on the denominator.

Hence $$0<S_s\le\frac{8100}{11}$$