A question in inequalities involving modulus

Question

How do I solve this sum:

Solve for $x \in \mathbb {R} $

$|x^2-2x|+|x-4|>|x^2-3x+4|$

Answer 1

By the triangle inequality, you have $\forall \, a,b \in \mathbb{R}$: $$|a+b| \le |a| + |b|$$ where the equality only holds when $a$ and $b$ have the same sign (both positive or both negative) or when at least one of $a$ and $b$ is equal to $0$; this means:

• $|a+b| = |a| + |b| \iff ab \ge 0$;
• $|a+b| < |a| + |b| \iff ab < 0$.

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Now notice that with $a=x^2-2x$ and $b=4-x$, you can rewrite: $$\left| x^2-3x+4 \right|=\left| x^2-2x+ (4-x) \right| \le \left| x^2-2x \right|+\left|4-x\right|$$ So all $x \in \mathbb{R}$ satisfy "$\le$" and for your "$<$", you only need to find out where the strict inequality holds and from above, we know this happens when $a$ and $b$ are non-zero and have a different sign, i.e. when $ab < 0$; so that leaves you with solving: $$\left( x^2-2x \right)\left(4-x\right) < 0 \iff x\left( x-2 \right)\left(4-x\right) < 0 \iff \ldots$$

Answer 2

Note that $|x^2-3x+4|=x^2-3x+4$ for all $x\in\mathbb{R}$ as $x^2-3x+4$ is always positive.

When $x\le 0$, $|x^2-2x|=x^2-2x$ and $|x-4|=4-x$. So

\begin{align*} x^2-2x+4-x&>x^2-3x+4\\ \end{align*}

which is impossible.

When $0<x\le 2$, $|x^2-2x|=2x-x^2$ and $|x-4|=4-x$. So

\begin{align*} 2x-x^2+4-x&>x^2-3x+4\\ 2x^2-4x&<0\\ 0<x&<2 \end{align*}

The solution for this case is $0<x<2$.

When $x>2$, $|x^2-2x|=x^2-2x$ and $|x-4|=x-4$. So

\begin{align*} x^2-2x+x-4&>x^2-3x+4\\ 2x&>8\\ x&>4 \end{align*} The solution in this case is $x>4$.

The final answer is $0<x<2$ or $x>4$.

Alternatively, we have $|x^2-2x|+|x-4|=|x^2-2x|+|4-x|\ge|(x^2-2x)+(4-x)|=|x^2-3x+4|$ by triangle inequality, with the equality holds if and only if $x^2-2x$ and $x-4$ have the same sign, i.e. $x(x-2)(4-x)\ge0$.

So $|x^2-2x|+|x-4|>|x^2-3x+4|$ if and only if $x(x-2)(4-x)<0$, i.e. $0<x<2$ or $x>4$.

Answer 3

Since for all reals $a$ and $b$ we have $$|a|+|b|=|a|+|-b|\geq|a-b|$$ and $x^2-2x-(x-4)=x^2-3x+4>0$, we get the answer: $\mathbb R\setminus$ $\{$cases of equality occuring$\}$

and since for $x\in(-\infty,0]$ or for $x\in[2,4]$ we get equality,

we obtain $$\mathbb R\setminus\left((-\infty,0]\cup[2,4]\right)=(0,2)\cup(4,+\infty)$$