Htykuut

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.

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

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

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*}

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:

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

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 $ale ble c$.




• 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 $bge 0$.




• 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
  $$
  begin{aligned}
  S
  &= frac{|a|}t +
  frac{|a+s|}{s+t} +
  frac{|a+s+t|}s
  \
  &ge
  frac{0}t +
  frac{s}{s+t} +
  frac{s+t}s
  %ge 2sqrt{
  %frac{s}{s+t} cdot
  %frac{s+t}s}
  ge2 ,qquad(age 0)
  end{aligned}
  $$
  and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.




• So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
  $$
  frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
  $$
  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:
  $$
  frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
  $$
  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 $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)