Good day I've tried today to solve the following exercise :

Prove that for all $a,b,c \in \mathbb{R}$ the following inequality:

$$ |a|+|b|+|c|+|a+b+c|-|a+b|-|a+c|+|b+c| \geq 0 $$

I've tried to use the triangle inequality, but I couldn't solve it

someone can give me a hint to solve it?

Best regards.

  • $\begingroup$ is $+|b+c|$ a typo? should it be $-|b+c|$ $\endgroup$ – chenbai Mar 30 '14 at 7:09
  • $\begingroup$ math.stackexchange.com/questions/591078/… $\endgroup$ – user94270 Mar 30 '14 at 7:58
  • $\begingroup$ Just use some triangular inequalities and you'll have your result. $\endgroup$ – Hakim Mar 30 '14 at 17:18

I think the right one is: $|a|+|b|+|c|+|a+b+c|-|a+b|-|a+c|-|b+c| \geq 0$

for $a,b,c$, there at least two of them have same sign ,WLOG, $ab \ge0 \implies |a|+|b|-|a+b|=0$

so it remains :$|c|+|a+b+c|\ge |b+c|+|a+c|$

it is trivial that if $c$ has same sign of $a$, then LHS=RHS.

if $ac<0$, WLOG,$a>b>0,c<0, d=-c>0 \iff |d|+|a+b-d|\ge |b-d|+|a-d|$

there is 4 cases:

case 1:$d<b$, LHS$=a+b$, RHS$=a+b-2d \implies LHS>RHS$

case2: $b<d<a$, LHS$=a+b$,RHS$=a-d+d-b=a-b \implies LHS>RHS$

case3: $a<d<a+b$, LHS$=a+b$,RHS$=d-a+d-b=d+d-(a+b)<d<a+b=LHS$

case 4:$d>a+b$, LHS$=2d-(a+b)$, RHS$=2d-(a+b)=$LHS




The triangle inequality reads $$|x+y|\leq |x|+|y|\tag{1}$$ and a standard proof of the reversed triangle inequality $$|x-y|\geq|x|-|y|\tag{2}$$ goes like this: "By (1) we have $|x|=|x-y+y|\leq|x-y|+|y|$, and thus (2) follow.".

Now, by (1) it is easy to see (do it!) $$2|a|+|b|+|c|-|a+b|-|a+c|\geq0$$ hence the inequality our main inequality would follow if we could prove $$|a|\leq|a+b+c|+|b+c|$$ (Why?)


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