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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.