# Nested absolute-value inequality

I try to solve a problem in two ways, but the results are not the same.

Method 1.

$$\lvert \lvert x \rvert + x \rvert \le 2$$ For $x < 0$, we have $\lvert x \rvert = -x$. Therefore:

$$\lvert -x+x \rvert \le 2$$

which is always true.

For $x \ge 0$, we have $\lvert x \rvert = x$. Therefore:

\begin{align} \lvert x + x \rvert & \le 2 \\ x & \le 1 \end{align}

Combining both answers, we get $x \le 1$.

Method 2.

$$\lvert \lvert x \rvert + x \rvert \le 2$$

Squaring both sides, we get:

\begin{align} (\lvert x \rvert + x)^2 & \le 4 \\ \lvert x \rvert^2 + 2x\lvert x \rvert + x^2 & \le 4 \\ 2x^2 + 2x\lvert x \rvert & \le 4 \\ x\lvert x \rvert & \le 2-x^2 \\ \end{align}

Squaring both sides again:

\begin{align} x^2 \lvert x \rvert^2 & \le 4 - 4x^2 + x^4 \\ x^4 & \le 4 - 4x^2 + x^4 \\ 0 & \le 4 - 4x^2 \\ x^2 & \le 1 \\ -1 & \le x \le 1 \end{align}

The answers are different, but I believe the first one is correct because if we substitute $x=-10$, for instance, we get the correct result. Where did I do wrong?

In method 2, the second time you square both sides, you have no guarantee that the sides you square will both be positive. Therefore the squaring will not necessarily preserve the inequality.

Solution $1$ is correct.

Solution $2$ is incorrect, the mistake is in the step:

$x|x|\leq 2-x^2 \iff x^2|x|^2\leq 4-4x^2+x^4$.

Clearly the left hand side holds for any negative value of $x$, the right side does not.

Squaring cannot eliminate nested absolute inequality. \begin{align} (|x| + x)^2 \\ |x| + 2x|x| + x^2 \\ 2x^2 + 2x|x| \end{align} And if you move $x$ to the other side of inequality, you should make it sign before squaring it in order to maintain the inequality. or instead of squaring it again:

For $x \lt 0$ \begin{align} x(-x) &\le 2 - x^2 \\ -x^2 &\le 2 - x^2 \\ 0 &\le 2 \tag{always true for $x \lt 0$} \end{align}

For $x \ge 0$ \begin{align} x(x) &\le 2 - x^2 \\ x^2 &\le 2 - x^2 \\ 2x^2 &\le 2 \\ x^2 &\le 1 \\ -1 \le x &\le 1 \end{align} Thus, combining both answer $x \le 1$

Squaring also eliminate discontinuity.
See graphic for $|x + |x||$ and compare with graphic for $(|x + |x||)^2$

• AFAIK, both functions are continuous? – guston76 Jan 13 '16 at 2:37
• You are right! more smooth? – inyoot Jan 13 '16 at 3:29