380 Algebra
the expression. Repeating, we can decrease the expression to one in which all numbers
are equal to^1 n. The value of the latter expression is(n+ 1 )n. This concludes the proof.
132.Project orthogonally the ellipse onto a plane to make it a circle. Because all areas
are multiplied by the same constant, namely the cosine of the angle made by the plane
of the ellipse and that of the projection, the problem translates to finding the largest area
triangles inscribed in a given circle. We apply Sturm’s principle, after we guess that all
these triangles have to be equilateral.
Starting with a triangle that is not equilateral, two cases can be distinguished. Either
the triangle is obtuse, in which case it lies inside a semidisk. Then its area is less than half
the area of the disk, and consequently smaller than the area of the inscribed equilateral
triangle. Or otherwise the triangle is acute. This is the case to which we apply the
principle.
Figure 61One of the sides of the triangle is larger than the side of the equilateral triangle and
one is smaller (since some side must subtend an arc greater than^23 πand another an arc
smaller than^23 π). Moving the vertex on the circle in the direction of the longer side
increases the area, as seen in Figure 61. We stop when one of the two sides becomes
equal to the side of the equilateral triangle. Repeating the procedure for the other two
sides, we eventually reach an equilateral triangle. In the process we kept increasing
the area. Therefore, the inscribed triangles that maximize the area are the equilateral
triangles (this method also proves that the maximum exists). These triangles are exactly
those whose centroid coincides with the center of the circle. Returning to the ellipse,
since the orthogonal projection preserves centroids, we conclude that the maximal-area
triangles inscribed an ellipse are those with the centroid at the center of the ellipse.
Remark.This last argument can be applied mutatis mutandis to show that of alln-gons
inscribed in a certain circle, the regular one has the largest area.
(12th W.L. Putnam Mathematical Competition, 1952)
133.The first inequality follows easily fromab≥abcandbc≥abc. For the second,
defineE(a, b, c)=ab+bc+ac− 2 abc. Assume thata≤b≤c,a<c, and let
α=min(^13 −a, c−^13 ), which is a positive number. We compute