I want to solve this inequality $$x^3-3 x^2+2 \sqrt{(x+2)^3}-6 x\geqslant 0.$$ I tried. Put $t = \sqrt{x + 2}$. Then, we get $$t^6+2t^3-9t^4+18t^2-8\geqslant 0.$$ Equavalent to $$(-2 + t)^2 (1 + t)^2 (-2 + 2 t + t^2)\geqslant 0.$$ From this inequality, we have $x\geqslant 2 \left(1-\sqrt{3}\right).$ How to solve this inequality with another way?
$\begingroup$
$\endgroup$
1
-
3$\begingroup$ Looks fine to me. Why do you want another proof? Is there something in your proof that makes you doubt? $\endgroup$– vonbrandMar 10, 2013 at 17:19
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As mentioned in the comment, your approach looks fine. Specifically, to approach a solution you need to "remove" the surd, and noting that:
$$ \sqrt{(x+2)^3}= \big({\sqrt{x+2}}\big)^3 $$
you're left with $ t = \sqrt{x+2} $ as the only useful substitution.
I should add: your "Equavalent to" step is really a big one, but if you showed that working, I see nothing to change.
