There are vertices, so we multiply this by ; since each edge, however, meets two vertices, every edge is counted twice. Therefore, we divide by  and conclude that there are edges. There are equally many edges of each length and lengths; hence there are edges of each length and the total of all edge lengths is . On the other hand,  is the total length of edges connected to a vertex on an -dimensional cube of equal volume, since in this case . Since the inequality says  it can be restated by multiplying through by to obtain  with equality if and only if   Thus the AM–GM inequality states that only the -cube has the smallest sum of lengths of edges connected to each vertex amongst all -dimensional boxes with the same volume. Consider the function  for all positive real numbers , and . Suppose we wish to find the minimal value of this function. First we rewrite it a bit: with  Applying the AM–GM inequality for , we get  Further, we know that the two sides are equal exactly when all the terms of the mean are equal:  All the points satisfying these conditions lie on a half-line starting at the origin and are given by  An important practical application in financial mathematics is to computing the rate of return: the annualized return, computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect. Jensen's inequality states that the value of a concave function of an arithmetic mean is greater than or equal to the arithmetic mean of the function's values. Since the logarithm function is concave, we have  Taking antilogs of the far left and far right sides, we have the AM–GM inequality. We have to show that  with equality only when all numbers are equal. If , then replacing both and by  Thus the right-hand side will be largest when all s are equal to the arithmetic mean  thus as this is then the largest value of right-hand side of the expression, we have  This is a valid proof for the case , but the procedure of taking iteratively pairwise averages may fail to produce equal numbers in the case . An example of this case is : Averaging two different numbers produces two equal numbers, but the third one is still different. Therefore, we never actually get an inequality involving the geometric mean of three equal numbers.