Intuition behind a certain proof of this inequality: $|a|+|b|+|c|-|a+b|-|a+c|-|b+c|+|a+b+c| \geq 0$

Question

One textbook exercise asks to prove $$|a|+|b|+|c|-|a+b|-|a+c|-|b+c|+|a+b+c| \geq 0.$$

The textbook's solution is:

If $a$, $b$ or $c$ is zero, the equality follows. Then, we can assume $|a| \geq |b| \geq |c| > 0$.

Dividing by $|a|$, the inequality is equivalent to

$$ 1 + |\frac{b}{a}| + |\frac{c}{a}| - |1+\frac{b}{a}| - |\frac{b}{a}+\frac{c}{a}| - |1+\frac{c}{a}| + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 $$

Since $ \frac{b}{a} \leq 1$ and $\frac{c}{a} \leq 1$, we can deduce that $|1+\frac{b}{a}| = 1+\frac{b}{a}$ and $|1+\frac{c}{a}| = > 1+\frac{c}{a}$.

Thus, it is sufficient to prove that

$$ |\frac{b}{a}| + |\frac{c}{a}| - |\frac{b}{a}+\frac{c}{a}| - (1+\frac{b}{a}+\frac{c}{a}) + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 .$$

Now, use the triangle inequality shows that the sum of the first three terms are positive, and absolute value shows that the sum of last two terms is also positive.

There may be more intuitive proofs to this, but how can one in 'some semi-logical way' arrive at this exact one?

Answer 1

Yes, by the triangle inequality $$\left|\frac{b}{a}\right|+\left|\frac{c}{a}\right|\geq\left|\frac{b}{a}+\frac{c}{a}\right|$$ and $$\left|1+\frac{b}{a}+\frac{c}{a}\right|\geq1+\frac{b}{a}+\frac{c}{a}$$ and after summing of these inequalities we are done.

The intuition behind this proof is very smooth I think: we want to delete as much as possible absolute values. We can see that it helps.

I think an easiest proof of this inequality it's a proof by using Popovici because $f(x)=|x|$ is a convex function. It's all proof!

Here is another example, how we can delete a symmetry or cyclicity, but we can prove an inequality.

We'll prove the following Schur's inequality.

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b+3abc\geq0.$$ Proof.

Let $a\geq b\geq c$.

Now, we'll delete the symmetry, but we'll prove the inequality.

$$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-c)(b-c)=$$ $$=(a-b)(a(a-c)-b(b-c))=(a-b)^2(a+b-c)\geq0.$$ Done!

Answer 2

Simply add $|a|+|b|\ge |a+b|$ etc. to arrive at $$\tag12|a|+2|b|+2|c|\ge |a+b|+|b+c|+|a+c|.$$ Also add $|a+b+c|\ge |a+b|+|c|$ etc. to find $$\tag2 3|a+b+c|\ge |a+b|+|a+c|+|b+c|+|a|+|b|+|c|$$ Combining $u(1)+v(2)$ with $u,v\ge 0$ and rearranging gives $$(2u-v)(|a|+|b|+|c|)-(u+v)(|a+b|+|b+c|+|a+c|)+3v|a+b+c|\ge0.$$ With $v=\frac13$, $u=\frac23$, we obtain the desired result.