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.
Answer
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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