# Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$

I would like to find the solution of

$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$

My try:

First I used the hint of this answer.

$$\frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$

Then the solution can be found by

$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$

But I think this is not the best approach.

• Several answers point out there is no solution. This is one reason why it is important, when presenting a problem like this, to include an explanation of why you believe the problem is correct - why do you think there is a solution? That explanation would typically come from the way that the inequality was derived, which is hard to see because this post doesn't include any source or motivation for the inequality. Commented Sep 25, 2018 at 20:38

First $$x^2 \le 8$$. Second, continuing from what you have written, i.e., $$\frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x,$$ we have $$\frac{-17}{\sqrt{8-0^2}+\sqrt{25-0^2}} \ge \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}} \ge x \implies x \le \frac{-17}{\sqrt{8}+5} \implies x^2 > 8.$$ Contradiction! So no real solutions.

• Nice.......(+1). Commented Sep 24, 2018 at 23:32
• I would like to accept two answers :( Thank you a lot!
– user596786
Commented Sep 24, 2018 at 23:42

Unfortunately the approach I did there is not useful here. However, here is another approach:

Write the inequality as

$$\sqrt{8-x^2}-x\geq\sqrt{25-x^2}. \tag{1}$$

Now use the Cauchy–Bunyakovsky–Schwarz inequality,

$$(x_1y_1+x_2y_2+\cdots+x_ny_n)^2\leq\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(y_1^2+y_2^2+\cdots+y_n^2\right),$$

to evaluate the left-side of $$(1)$$. Using this formula, one gets

$$\left(\sqrt{8-x^2}-x\right)^2\leq\left(1+1\right)\left(8-x^2+x^2\right)=16.$$

Notice we must have $$-2\sqrt{2}\leq x \leq 2\sqrt{2}$$ since, see Doug M's answer, $$>$$ and $$<$$ are not defined concepts over the complex numbers (sorry for that). Then, the right-side of $$(1)$$ takes

$$\sqrt{25-x^2}\geq \sqrt{25-8}=\sqrt{17}.$$

Futher, notice also that the left-side of $$(1)$$ is less than or equal to $$4$$, and that the right-side is greater than or equal to $$4$$. Therefore, there's no real solution.

• I noticed that the votes are not increasing my reputation. :( Commented Sep 24, 2018 at 23:06
• That's probably something you should mention on the meta site (if there's not an FAQ about it). Commented Sep 24, 2018 at 23:16
• Hi, the max reputation per day is $200$, I think it can go past that limit if you get your answer marked as accepted or something that isnt an up-vote. Commented Sep 24, 2018 at 23:24
• @Shaun and Dahaka Thank you. Commented Sep 24, 2018 at 23:33
• @DineshShankar Thank you for this another approach.
– user596786
Commented Sep 24, 2018 at 23:39

Since $$>,<$$ are not defined concepts over the complex numbers, we can assume that $$\sqrt{8-x^2}, \sqrt{25-x^2}$$ are real

And $$\sqrt{25-x^2} > \sqrt{8-x^2}$$

Which implies $$x\in [-2\sqrt 2,0]$$

but for all $$x$$ in this interval, $$\sqrt{8-x^2} - \sqrt{25-x^2} > x$$

there is no solution.

Dinesh's answer is alright, but there is another approach. Your question can be rephrased as finding the range of $$x$$ such that $$f(x)\geq 0$$, where $$f(x) = \sqrt{8-x^2}-\sqrt{25-x^2}-x.$$ We find the derivative of $$f$$ then set it to be $$0$$ to find its extreme values: $$f'(x) = -\frac{x}{\sqrt{8-x^2}}+\frac{x}{\sqrt{25-x^2}}-1=0$$ Which implies that $$x\left(\sqrt{8-x^2}-\sqrt{25-x^2}\right)=\sqrt{(8-x^2)(25-x^2)}.$$ Solve this (with some effort) to get that the critical point of $$f$$, which happens to be a maximum, occurs at about $$x=-2.37$$ where the $$y$$ value is $$-0.49<0$$. Thus $$f$$ is always less than $$0$$ and there is no solution.