$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$

Question

If $a, b, c$ are distinct real numbers then you demonstrate that:

$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$

Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$

For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.

Answer 1

Let $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$.

Thus, $$xy+xz+yz=\sum_{cyc}\frac{ab}{(b-c)(c-a)}=\frac{\sum\limits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$ Id est, $$\sum_{cyc}\left|\frac{a}{b-c}\right|=\sqrt{\left(|x|+|y|+|z|\right)^2}=\sqrt{x^2+y^2+z^2+2\sum\limits_{cyc}|xy|}=$$ $$=\sqrt{(x+y+z)^2+2+2\sum\limits_{cyc}|xy|}\geq\sqrt{2+2}=2.$$

Answer 2

Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|\le |c|+|a|$ and $|a-b| \le |a|+|b-c|+|c|$. Therefore $$ \frac{|a|}{|b-c|}+\frac{|b|}{|c-a|}+\frac{|c|}{|a-b|} \ge \frac{|a|}{|b-c|} + \frac{|b-c|+|c|}{|c|+|a|} + \frac{|c|}{|a|+|b-c|+|c|} = (*).$$

Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 \ge 2xy$, we get \begin{align*} (*) & = \frac{x}{y}+\frac{y+z}{z+x}+\frac{z}{x+y+z} \\ &\ge \frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \\ &= \frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\ &\ge \frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\ &=2.\end{align*}

Answer 3

Notice that $|x-y|\leq |x|+|y|$ implies that: $$\frac{|z|}{|x-y|} \geq \frac{|z|}{|x|+|y|}.$$ Then:

$$S \geq \frac{|a|}{|b|+|c|} + \frac{|b|}{|a|+|c|} + \frac{|c|}{|a|+|b|} = \\ =\frac{A}{B+C} + \frac{B}{A+C} + \frac{C}{A+B},$$ where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.

Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:

$$S \geq \frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$

Notice that the denominator can be expanded as follows:

$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$

Therefore, the numerator can be rewritten as:

$$ N= D + A^3 +B^3 + C^3 + ABC.$$

In other words:

$$S \geq \frac{D + A^3 +B^3 + C^3 + ABC}{D} > \frac{D}{D} = 1.$$

Answer 4

I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.

• First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $a\le b\le c$.

• Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.

• Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $b\ge 0$.

• The given inequality can now be rewritten equivalently: $$ S= \frac{|a|}t + \frac{|a+s|}{s+t} + \frac{|a+s+t|}s \ge 2 \ . $$

• If $a$ is $\ge0$, then all three numbers $a,b,c$ are $\ge 0$, and we can write $$ \begin{aligned} S &= \frac{|a|}t + \frac{|a+s|}{s+t} + \frac{|a+s+t|}s \\ &\ge \frac{0}t + \frac{s}{s+t} + \frac{s+t}s %\ge 2\sqrt{ %\frac{s}{s+t} \cdot %\frac{s+t}s} \ge2\ ,\qquad(a\ge 0) \end{aligned} $$ and the last $\ge$ (AM-GM) is moreover $>0$ (because $t\ne 0$). The case with $a < b < c=a+s+t\le 0$ is similar.

• So the single interesting case is the one with $a<0\le b<c$ . We rewrite the inequality in the form: $$ \frac{|b-s|}t+\frac b{s+t}+\frac{b+t}s\ge 2\ . $$ Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show: $$ \frac{s-b}t+\frac b{s+t}+\frac{b+t}s\ge 2\ . $$ But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $\frac st+\frac ts\ge 2$, true, with equality for $s=t$. For $b=s$ we get $\frac s{s+t}+\frac {s+t}s\ge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)