Cyclic Inequality

Consider the solution set to the inequality $(b + 2c) - ( a+ 2b ) > 0$ . It corresponds to the "upper half" of the space that is cut by the plane $-a - b + 2c = 0$ .
Consider the solution set to the inequality $(c+2a) - (b+2c) > 0$ . It corresponds to the "upper half" of the space that is cut by the plane $2a - b -c = 0$ .

Because these 2 planes are not parallel, hence we can conclude that they must intersect, and so these "upper half" will have to intersect, giving us a solution to the inequality.

---

This is an example of a solution where we have shown that the answer exists, but did not have to find a value that worked.

The 2-variable version is slightly more interesting / relevant. If we have inequality $ax + by > c, dx + ey > f$ , then if $ae \neq bd$ , we can conclude that there will always be a solution (though it's not clear what the solution is in terms of the variables).

My goal is not only to find a "yes" (one example is enough, after all), but also to establish precise conditions for the inequalities to hold true.

First I subtracted $a+b+c$ . The inequality then becomes $b-c < c-a < a-b.$ Since the sum of these three terms is zero, it is obvious that at least one of them must be negative and one positive. This shows immediately that $b < a, c$ . The order of $a, c$ does not matter, as long as they are sufficiently close to each other.

Thus we have the following recipe to make all triples that satisfy the inequalities:

• pick any number $b$

• pick any number $c > b$

• pick any number $a$ between $\tfrac12(b+c)$ and $c$

• swap $a$ and $c$ if desired.

To prove that this works, let $c = b + k$ and $a = b + k t$ , with $k > 0$ and $0 < t < 1$ . Then the inequality above translates to $-k < k (1-t) < k t,$ which is true iff $\tfrac12 < t < 1$ .

Or, if we swap $a$ and $c$ , let $a = b + k$ and $c = b + k t$ , with $k > 0$ . Then the inequality becomes $-k t < -k (1-t) < k,$ which again is true iff $\tfrac12 < t < 1$ .

Comparing the three expressions pairwise, we have that

$a + 2b \lt b + 2c \Longrightarrow \dfrac{a + b}{2} \lt c$ , (i),

$b + 2c \lt c + 2a \Longrightarrow \dfrac{b + c}{2} \lt a$ , (ii),

So considering a number line, from (i) we have that the average of $a,b$ lies to the left of $c$ , so at least one of $a,b$ lies to the left of $c$ . From (ii) the average of $b,c$ lies to the left of $a$ , so at least one of $b,c$ lies to the left of $a$ . So neither of $a,c$ can be the least value, which implies that $b$ is less than both $a$ and $c$ . We can't glean any information from these inequalities as to how $a$ and $c$ relate to one another, so we will try plugging some examples of $(a,b,c)$ into the given inequality string with $b$ as the minimum to see what pans out:

• $(a,b,c) = (2,1,3) \Longrightarrow a + 2b = 4, b + 2c = 7, c + 2a = 7$ , inequality not satisfied;

• $(a,b,c) = (3,1,4) \Longrightarrow a + 2b = 5, b + 2c = 9, c + 2a = 10$ , inequality satisfied;

• $(a,b,c) = (4,1,3) \Longrightarrow a + 2b = 6, b + 2c = 7, c + 2a = 11$ , inequality satisfied.

So $\boxed{\text{Yes}}$ , there do exist distinct $a,b,c$ such that $a + 2b \lt b + 2c \lt c + 2a$ . As a necessary condition $b$ must be the least of the values, but this is not a sufficient condition, i.e. we can have $b$ as the minimum but choose $a,c$ so that the inequality is not satisfied. Also, we can find cases where either $a \lt c$ or $c \lt a$ and still satisfy the inequality, so there are no restrictions on the relative values of $a,c$ .

Comments: If, for example, $c = a + 1$ then $b + 2c \lt c + 2a \Longrightarrow b \lt a - 1$ , which explains why $(a,b,c) = (2,1,3)$ did not work. Similarly, if $a = c + 1$ then $a + 2b \lt b + 2c \Longrightarrow b \lt c - 1$ . So $a,c$ are structurally equivalent in the given inequality string, but what matters in finding triples $(a,b,c)$ that work is the absolute value of the difference between $a$ and $c$ . In general, if $|a - c| = k$ then we require that both $a$ and $c$ exceed $b + k$ .

a + 2b < b + 2c (i)
b +2c < c + 2a (ii)
a + 2b < c + 2a (iii)

Add them all together, we get: 2a + 2c +5b < 4c + 4a + b => 4b < 2c + 2a => [2b < c + a]

When we manipulate (iii), which connects the minimum value [a + 2b] and the max [c + 2a], we get [2b < c + a] as well, hence the inequality is possible.

a + 2b < b + 2c < c + 2a

Implies that

a + 2b < c + 2a (transitive quality)

hence,

2b < c + a

then,

b < $\frac{c + a}{2}$

is the required condition.

e.g.

a = 10 , b = 3 and c = 8 When substituted, gives the 'chain'

16 < 19 < 28

as required.

regards,

L3mon3ntry.

We can take two examples, in which we consider rational numbers:

$1$ ) $a=2'5$ , $b=1$ and $c=2$ , and the double inequality is: $4,5<5<7$ , so it is True;

$2$ ) $a=1$ , $b=2$ and $c=3$ , and the double inequality is: $5<8<5$ , so it is False.

Then, you have that the double inequality is not True for any number of rational numbers, worse if there are triples for which it is.

Consider the numbers $(a, b, c) = (7, 4, 8)$ .