Solving an inequality in 2 similar ways gives opposite solution

Question

I have the following inequality: $x+\sqrt{2x^2-4x+4} \ge 0$ Here are the 2 ways how I tried to solve it:

No. 1:

$x+\sqrt{2x^2-4x+4} \ge 0$

$x \ge -\sqrt{2x^2-4x+4}$

$x^2 \ge 2x^2-4x+4$

$0 \ge x^2-4x+4$

$(x-2)^2 \le 0$

The last inequality doesn't make sense, so I tried the other way.

No. 2:

$x+\sqrt{2x^2-4x+4} \ge 0$

$\sqrt{2x^2-4x+4} \ge -x$

$2x^2-4x+4 \ge x^2$

$x^2-4x+4 \ge 0$

$(x-2)^2 \ge 0$

I cannot figure out where did I make the mistake in the first try.

Answer 1

Hint: Note that $$x\geq y\not\Rightarrow x^2\geq y^2$$ For instance: $$1>-2\Rightarrow1<4$$ Also, note that $$\sqrt{x}\geq0,\ \forall\ x\geq0$$ Can you figure out your mistake?

Answer 2

No. 2:

$x+\sqrt{2x^2-4x+4} \ge 0$

$\sqrt{2x^2-4x+4} \ge -x$

$2x^2-4x+4 \ge x^2$

$x^2-4x+4 \ge 0$

$(x-2)^2 \ge 0$

Here, $\sqrt{2x^2-4x+4} \ge -x$, since LHS is positive, and RHS can be negative or positive. Hence out final solution would be $(x-2)^2 \ge 0$. This means x ∈ R .

Answer 3

First of all, you must determine for which $x$ an inequality makes sense. Here the condition is $$2x^2-4x+4=2\bigl((x-1)^2+1\bigr)\ge 0,$$ so it makes sense for all $x$. You can rewrite it as $$\sqrt{2x^2-4x+4}\ge -x.$$ Now remember that, if $A\ge 0$, $$\sqrt A\ge B\iff A\ge B^2\quad\text{OR}\quad B\le 0$$ so here it is equivalent to $$2x^2-4x+4\ge x^2\quad\text{OR}\quad x\ge 0\iff x^2-4x+4=(x-2)^2\ge0\quad \text{OR}\quad x\ge 0. $$ As the first term of this disjunction is always true, the given inequality is always true.

Answer 4

Alternative way: $\;\sqrt{2x^2-4x+4} = \sqrt{x^2+(x-2)^2} \ge \sqrt{x^2} = |x| \ge -x\,$ for all $\,\forall x \in \mathbb{R}\,$.