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I have the following problem in an assignment and have been struggling to do it.

$2 + 2x - x^2 \geq 2 \sqrt{1+2x}$

I have tried solving for $x$ but have not been able to do so. Any hints to solve this inequality would be greatly appreciated. Thanks in advance.

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Hints:

$2 + 2x - x^2 \geq 2 \sqrt{1+2x}$

$(1+2x)-2\sqrt{1+2x}+1-x^2\geq 0 $

$(\sqrt{1+2x}-1)^2-x^2\geq 0$

Can you make it from here?

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Essentially, we have the following inequality:

$2 + 2x - x^2 \ge 2\sqrt{1 + 2x}$

$\Rightarrow x^2 - 2x - 2 \le -2\sqrt{1 + 2x} $

The right-hand-side is negative, because the square-root will be positive. That tells us that the quadratic will also have to be negative. In which interval is it negative? Between the two roots, i.e., in the interval $\left[1 - \sqrt{3}, 1 + \sqrt 3\right]$. But $x \ge -\frac{1}{2}$, which restricts our interval to $\left[-\frac{1}{2},1 + \sqrt3\right]$.

Now that we've got the interval within which $x$ must fall, let us proceed with squaring which can be done when both the sides are positive.

$\Rightarrow 2x + 2 -x^2 \ge 2\sqrt{1+2x}$

$\Rightarrow \left(2x + 2 - x^2\right)^2 \ge 4 + 8x$

$\Rightarrow 4x^2+ 8x + 4 + x^4 - 4x^2(x+1) \ge 4+8x$

$\Rightarrow x^4 - 4x^3 \ge 0$

$\Rightarrow x^3(x - 4) \ge 0$

$\Rightarrow x \ge 4 ~ {\rm or} ~ x \le 0.$

But if we look at the restricted interval, we have the answer $\boxed{x \in \left[-\frac{1}{2},0 \right]}$.

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