For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$

Question

For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$

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Solution: $$ |a| + |b| + |c| \geq |b+c| + |c+a| + |a+b| \geq 0 $$ and $$0 \leq |a+b+c|\leq |a+b| + |c| \leq |a| + |b| + |c|$$ so we have $$0 \leq |a+b+c| \geq 0 $$ $$ |a+b+c|= 0 \iff a+b+c =0 $$

I need check my solution. Many thanks!

Answer 1

hint: Square each of the $| ..|$ inequality and consider $f(a) = a^2 + (2b+2c)a + (b+c)^2$. Show that $f(a) \ge 0$.

Answer 2

We have: $$a^2\geq(b+c)^2,$$ $$b^2\geq(a+c)^2$$ and $$c^2\geq(a+b)^2$$ or $$(a-b-c)(a+b+c)\geq0,$$ $$(b-a-c)(a+b+c)\geq0$$ and $$(c-a-b)(a+b+c)\geq0.$$ Now, for $a+b+c>0$ we obtain: $$a-b-c\geq0,$$ $$b-a-c\geq0$$ and $$c-a-b\geq0,$$ which gives $$a-b-c+b-a-c\geq0$$ or $$c\leq0.$$ Similarly, $b\leq0$ and $a\leq0,$ which gives $$a+b+c\leq0,$$ which is a contradiction.

By the same way we'll obtain a contradiction for $a+b+c<0.$

Thus, $a+b+c=0$ and we are done because for $a=b=c=0$ we obtain $a+b+c=0.$

Answer 3

For a real number $x$ we have $$ |x| \ge |1-x| \iff x \ge \frac 12 \, . $$ This can easily be seen by squaring the left-hand side, or geometrically: Which real numbers are closer to $1$ than to zero?

Now assume that $s = a+b+c$ is not zero. Then $$ |a| \ge |b+c| \iff |a| \ge |s-a| \iff \left| \frac as \right| \ge \left| 1 - \frac as \right| \iff \frac as \ge \frac 12 \, . $$ Similarly, $\frac bs \ge \frac 12$ and $\frac cs \ge \frac 12$, and adding these inequalities gives a contradiction: $$ 1 = \frac{a+b+c}{s} \ge \frac 32 \, . $$

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The nice thing about this approach is that it works with complex numbers $a, b, c \in \Bbb C$ as well. The location of all complex numbers which are closer to one than to zero (or at the same distance) is a half-plane: $$ |z| \ge |1-z| \iff \operatorname{Re} z \ge \frac 12 \, . $$

Then $$ |a| \ge |b+c| \iff |a| \ge |s-a| \iff \left| \frac as \right| \ge \left| 1 - \frac as \right| \iff \operatorname{Re}\left(\frac as \right) \ge \frac 12 $$ and adding these estimates for $a, b, c$ again gives a contradiction.