Help with the inequality $a+{k}r\leq c+\sqrt{k}s$

Question

Let $k\geq 0$. I want to find an upper bound for $k$ given that $$a+{k}r\leq c+\sqrt{k}s$$ with $a,c,r,s$ any real constants.

My attempt was to write $$(kr-\sqrt{k}s)^2\leq (c-a)^2$$ but can't go any further.

Any hint will be appreciated.

Answer 1

If $x=\sqrt{k}$ then you have to solve:

$$ rx^2 -sx +a-c \leq 0$$

If $x_1\leq x_2$ are solutions to the corresponding quadratic equation then we have:

• If $r>0$ then $x\in [x_1,x_2]\cap [0,\infty)$

So if $x_2\geq 0$ then $k_{\max} =x_2^2$ else it does not exists.

• If $r<0$ then $x\in \Big((-\infty, x_1]\cup [x_2,\infty)\Big) \cap [0,\infty)$

Maximal value for $k$ does not exists.

Answer 2

For now, pretend that the ≤ sign is an = sign. Then:

$$a+{k}r = c+\sqrt{k}s$$ $$a+kr-c = \sqrt{k}s$$ $$(a+kr-c)^2 = ks^2$$ $$a^2 + akr - ac + akr + k^2r^2 - ckr - ac - ckr + c^2 = ks^2$$ $$a^2 + 2akr - 2ac + k^2r^2 - 2ckr + c^2 - ks^2 = 0$$ $$r^2k^2 + (2ar - 2cr - s^2)k + (a^2 - 2ac + c^2) = 0$$

This gives you a quadratic in terms of $k$.

Recall that for a parabola expressed as $y = Ax^2 + Bx + C$, the vertex is at $x = \frac{-B}{2A}$ (and $y = \frac{4AC-B^2}{4A}$). This is a maximum if $A < 0$, or a minimum if $A > 0$.