Question

can someone solve this ?
|x+1|^2 + 2 |x+2| ≥ 2

Answer 1

we can note this:
\[\Large {\left| {x + 1} \right|^2} = {\left( {x + 1} \right)^2}\]

Answer 2

There's really no difference between \(|x+1|^2\) and \((x+1)^2\) since they both evaluate to the same thing, so you can safely throw these absolute value signs away

Answer 3

welp

Answer 4

so, we can rewrite your inequality, as below:
\[\Large {\left( {x + 1} \right)^2} + 2\left| {x + 2} \right| \geqslant 2\]

Answer 5

so @Michele_Laino (x+1)^2 = x^2+2x+1 ?

Answer 6

next, we have to consider the subsequent two cases:
\[\large \left\{ \begin{gathered}
{\left( {x + 1} \right)^2} + 2\left( {x + 2} \right) \geqslant 2 \hfill \\
x + 2 \geqslant 0 \hfill \\
\end{gathered} \right.,\quad \left\{ \begin{gathered}
{\left( {x + 1} \right)^2} - 2\left( {x + 2} \right) \geqslant 2 \hfill \\
x + 2 < 0 \hfill \\
\end{gathered} \right.\]
please solve both systems above

Answer 7

did you solve it