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We have the following inequality: $$ x < - \sqrt{x^2-8x+4} $$ It is very easy to check that it has no solution (graphically). But I'm having a lot of trouble with a proof. I know for instance that squaring both sides is no good because the $(-)$ sign disappears. Can anybody suggest a way to do it? Thanks.

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  • $\begingroup$ Often considering the cases where the LHS/RHS are positive/negative help when you want to square both sides. $\endgroup$ – Shuri2060 Jul 2 '17 at 1:47
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    $\begingroup$ " I know for instance that squaring both sides is no good because the - sign disappears." That's not a problem at all because you know x is negative. So square it and flip the inequality. $\endgroup$ – fleablood Jul 2 '17 at 4:06
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The inequality does not hold for non-negative $x$, because the left hand side is non-negative but the right hand side is non-positive.

For $x <0$, both sides of the inequality is non-positive. Square the original inequality and note that the inequality sign should be reversed,

$$\begin{align*} x &< -\sqrt{x^2-8x+4}\\ x^2 &> x^2 - 8x + 4\\ 0 &> -8x + 4\\ x &> \frac12 \end{align*}$$

The assumption was that $x < 0$, so there is also no solution for the case $x < 0$.

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We need to solve $$\sqrt{x^2-8x+4}<-x,$$ which gives $-x>0$ and $x^2-8x+4\geq0$.

Now we can use squaring of the both sides, which gives $$x^2-8x+4<x^2$$ or $$x>\frac{1}{2},$$ which is impossible because we have at least $-x>0$.

Thus, our inequality has no solutions.

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To prove $x < - \sqrt{x^2-8x+4}$ is not true for all $x \in \mathbb{R}$, you can instead prove that $x \ge - \sqrt{x^2-8x+4} \,\,\,\,\,\forall\,\,\,\,\, x \in \mathbb{R}$ when the RHS is defined as the two statements are equivalent using the law of trichotomy.

It is easy to see this holds for $x>0$ as the LHS is positive while the RHS is not.

Assume $x<0$.

Then

$$x \le \frac{1}{2}$$

$$\iff 0 \le -8x+4$$

$$\iff 0 \lt x^2 \le x^2-8x+4$$

$$\iff 0 \lt \sqrt{x^2} \le \sqrt{x^2-8x+4}$$

$(x<0)$

$$\iff -x \le \sqrt{x^2-8x+4}$$

$$\iff x \ge -\sqrt{x^2-8x+4}$$


This is essentially the same method as peterwhy's, but a different way of looking at it.

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