I need help with another inequality.


Squaring leads nowhere as you get a polynomial of degree 4 with no rational roots.

  • $\begingroup$ What about using a conjugate expression? $\endgroup$ – abiessu Sep 10 '14 at 18:55
  • $\begingroup$ I tried that too but it also leeds nowhere. I got a common multiple 3x - 2 but the remaining expression is of degree 4 after squaring. $\endgroup$ – chen h. Sep 10 '14 at 18:58

To start with,


immediately forces the bounds $x\in[-2,2].$ Using an LHS conjugate, we have

$$(\sqrt{2x+4}-2\sqrt{2-x})\frac {\sqrt{2x+4}+2\sqrt{2-x}}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac {6x-4}{\sqrt{2x+4}+2\sqrt{2-x}}\gt\frac{12x-8}{\sqrt{9x^2+16}}$$

Since $x\lt \dfrac 23$ causes $6x-4$ to be negative, evaluate

$$x\in\left[-2,\frac 23\right), \frac {1}{\sqrt{2x+4}+2\sqrt{2-x}}\lt\frac{2}{\sqrt{9x^2+16}}\tag 1$$ $$x\in\left(\frac 23,2\right], \frac {1}{\sqrt{2x+4}+2\sqrt{2-x}}\gt\frac{2}{\sqrt{9x^2+16}}\tag 2$$

(Note that $x=\dfrac 23$ causes the sides to be equal, and therefore the inequality is false at that point.)

Now change $(1)$ by fraction flip to

$$\sqrt{2x+4}+2\sqrt{2-x}\gt\sqrt{\frac 94x^2+4}\tag 3$$

and similarly for $(2)$:

$$\sqrt{2x+4}+2\sqrt{2-x}\lt\sqrt{\frac 94x^2+4}\tag 4$$

Squaring both sides at this point results in

$$\begin{align}2x+4+4-2x+4\sqrt{2(4-x^2)}\quad &R \quad \frac 94x^2+4\\ \sqrt{2(4-x^2)}\quad &R\quad \frac 9{16}x^2-1\end{align}$$

Now we see that the LHS is strictly non-negative over the full range, while the RHS is positive only for $|x|\gt \frac 34$. This means that $(4)$ is not valid unless $x\in[\frac 34, 2]$ as well as $(2)$ and therefore we need only consider $x\lt \frac 23$ or $x\gt\frac 34$, and in fact, the range $x\in[-\frac 34,\frac 23)$ is valid, and we only need to consider $x\in[-2,-\frac 34)\cup (\frac 34, 2]$ since we know that the LHS goes to $0$ at $x=-2,2$.

Squaring again will now achieve

$$\begin{align}8-2x^2&\gt\frac{81}{256}x^4-\frac{9}{8}x^2+1\\ 0&\gt 81x^4+7\cdot 32x^2-7\cdot 256\end{align}\tag 5$$

This is really just a quadratic in $x^2$, and need only be solved for a value $x_0\in[-2,-\frac 34)\cup [\frac 34, 2]$. Such a solution of $81x^4+7\cdot 32x^2-7\cdot 256=0$, produces the result that $x\in(-|x_0|,\frac 23)\cup (|x_0|,2)$ satisfies our original inequality. Note that since it is a quadratic in $x^2$ the range is (mostly) symmetric about $x=0$ with the exception of the $0$ at $x=\frac 23$ from the initial condition.

  • $\begingroup$ Thanks, I already reached that point. After getting rid of the denominators, the remaining inequality will be of degree 4 after squaring. :( $\endgroup$ – chen h. Sep 10 '14 at 19:14
  • $\begingroup$ See what you think of that degree-$4$ inequality. :-) $\endgroup$ – abiessu Sep 10 '14 at 19:35
  • $\begingroup$ Thanks. It looks like an IMO problem. Very hard! $\endgroup$ – chen h. Sep 10 '14 at 20:12
  • $\begingroup$ You're welcome. :-) $\endgroup$ – abiessu Sep 10 '14 at 20:16
  • $\begingroup$ Oops, I missed one bit of range for a solution $x_0\in[\frac 34, 2]$, I've updated to add that. $\endgroup$ – abiessu Sep 10 '14 at 20:25

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