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Proof of an

Michael Trott

Let , , be the lengths of the sides of a non-degenerate triangle. Then the following inequality holds (M. Bjelica. FILOMAT 9, 117 (1995)).

We will use Mathematica to prove this inequality in a straightforward way.

Without loss of generality we may assume . The left hand side then reduces to a function of two variables.

Let us look at a contour plot of this function to get a feel for this problem.

We take the gradient of to find a local minimum.

Since the triangle is non-degenerate, the denominators can never vanish. Therefore we can concentrate on the numerators.

To find all the roots of we will eliminate all the square roots to construct an equivalent polynomial. There are three types of square roots.

We introduce auxiliary variables for the square roots.

Eliminating the auxiliary variables , , and gives two polynomials.

The common zeros of these two polynomials include the zeros of . Here are their degrees.

All the powers are even, so we introduce new variables to halve the degrees.

Eliminating the variable from leaves a univariate polynomial in .

Eliminating the variable from leaves a univariate polynomial in .

These are the roots of for .

We select the positive ones.

These are the roots of for .

Again we select the positive ones.

So we arrive at the following 81 potential minimizing pairs for .

We delete the pairs that do not make the two functions of vanish.

We substitute these four pairs into the original expression and evaluate numerically.

This shows that the minimum value is , when the triangle is equilateral.

Finally, we must investigate the behaviour near the boundary of the region.

As tends to infinity, approaches 2. Since this is greater than , the local minimum we have found is the global minimum.

Formatting SamplerIntegrals, Sums and Products