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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$\begingroup$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 $x < 0$.
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$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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