Let $x \in R$. Then, prove that $x^2+|x-6|>5$

Question

I have tried to prove this statement by utilizing the proof by cases method. My cases are (1)$x=6$, (2)$x>6$ and (3)$x<6$.

For (3) for some reason it's not true

Case (1): For $x>6$, I know that $x^2+x>11$ is true

Case (2): For $x=6$, clearly $36>5$

Case (3):

For $x<6$, $x^2+x>11$ it is not true. (e.g 1,2,3...)

Thanks

Answer 1

For your case $(3)$, notice that if $x<6$ then $|x-6|=6-x$. Then, your inequality becomes $$x^2+6-x>5$$ $$x^2-x+1>0$$ $$(x-\frac{1}{2})^2+\frac{3}{4}>0$$ which is true.

Answer 2

Your attempt seems incorrect to me when you analyze the case $3$.

Note that, if $x≤6$, then you have:

$$ \begin{align} &x^2-x+6-5>0\wedge x≤6\\ \implies &\left(x-\frac 12\right)^2+\frac 34 >0. \end{align} $$