# Solution to puzzle 59: Triangle inequality

A triangle has sides of length a, b, and c.  Show that

## Left inequality

The left side of the inequality is, in fact, true for all triples (a, b, c) of positive real numbers.  We can prove it using the rearrangement inequality, stated below.

Let a1  a2  ...  an and b1  b2  ...  bn be real numbers.  For any permutation (c1, c2, ..., cn) of (b1, b2, ..., bn), we have:

a1b1 + a2b2 + ... + anbn    a1c1 + a2c2 + ... + ancn    a1bn + a2bn−1 + ... + anb1,

with equality if, and only if, (c1, c2, ..., cn) is equal to (b1, b2, ..., bn) or (bn, bn−1, ..., b1), respectively.

That is, the sum is maximal when the two sequences, {ai} and {bi}, are sorted in the same way, and is minimal when they are sorted oppositely.

Now we apply the rearrangement inequality to suitably chosen sequences.  Specifically, we will use the result that the sum is maximal when the two sequences are sorted in the same way.

Without loss of generality, assume a b c.
Then the sequences {a, b, c} and    are sorted the same way.

We twice rotate the second sequence, and apply the rearrangement inequality, to obtain:

Adding these two inequalities, and dividing by two, we get

We must also show that equality can occur, which is readily seen by setting a = b = c.

## Right inequality

In order to prove the right inequality, we must use the fact that a, b, c are the sides of a triangle.

Let s = ½(a + b + c) be the semi-perimeter of the triangle.
In any triangle, a + b > c, and so a + b > s.

Hence     and similarly

Adding the three inequalities, we get

Therefore

The following inequality is due to Gheorge Eckstein.

Let a, b, x, y, z be positive real numbers.  Show that: