Proving Algebraic Inequalities
If x and y are positive real numbers then clearly the sum x/y + y/x is greater than or equal to 2, because the quantity x/y + y/x - 2 can be multiplied through by xy (which is positive if x and y are both positive) to give x2 + y2 - 2xy, which can also be written as (x-y)2. Since the square of a real number is always non-negative, this proves that x/y + y/x is never less than 2. Of course this proof is also valid if x and y are both negative real numbers, but not if one is positive and the other negative, because in that case the inequality is reversed when we multiply by xy.
Now suppose we want to prove the slightly more complicated inequality
for any positive real values of x, y, and z. There are several ways we could approach this. For example, if we define k = x + y + z then the quantity on the left hand side can be written as
Dividing all the numerators and denominators by k, and defining the positive variables X = x/k, Y = y/k, and Z = z/k, we have
We note that X + Y + Z = 1, and each of the numbers X,Y,Z is less than 1. Hence each term is a convergent geometric series, e.g., the first term equals the infinite sum X + X2 + X3 + ... Combining terms of equal powers gives
The first quantity in parentheses is equal to 1, whereas the remaining terms very depending on the partition of 1 into the three positive summands X, Y, and Z. It's intuitively obvious that each sum of powers is minimized by the equal partition, i.e., by setting X = Y = Z = 1/3. To show this rigorously, we can substitute 1- X- Y in place of Z and then set the partial derivatives of the general sum of powers with respect to X and Y equal to 0. The derivative with respect to X gives
which implies X = 1- X- Y = Z. Likewise the derivative with respect to Y implies Y = Z. Substituting 1/3 for X, Y, and Z into the prior expression gives 3/2 for the minimum possible value, which is achieved when, and only when, x = y = z.
This gives a reasonably straightforward proof, and it immediately generalizes to the case of n variables. For example, we have
for four positive variables, and in general for n variables the sum is no less than n/(n-1). However, strictly speaking the proof involves calculus, whereas in the case of just two variables the inequality x/y + y/x 2 is due to the simple algebraic fact that (x- y)2 cannot be negative. It's tempting to think that every simple algebraic inequality (at least those that permit equality) must be expressible in the form of an algebraic square (or sum of squares) being non-negative. In the case of three variables, it is possible to give a purely algebraic proof on this basis?
Notice that in the case of two variables we had equality x/y + y/x = 2 if and only if x = y. This is consistent with the fact that the quantity (x-y)2 vanishes if and only if x = y. Similarly in the case of three variables we have equality if and only if x = y = z, and this suggests that if we are to express the inequality in terms of a square (or sum of squares) being non-negative, the quantities to be squared must vanish when x = y = z. Hence we expect the solution must involve the quantities (x-y), (x-z), and (y-z). Unfortunately, the sum of the squares of these quantities is not a factor of the desired inequality.
However, we notice that the vanishing of the sum of squares of those three differences is equivalent to the vanishing of the sum of squares of any two of them (by transitivity). To give the correct degree and maintain symmetry we are then led to the desired algebraic identity
Since x,y,z are positive, this quantity is necessarily non-negative, and vanishes if and only if x = y = z. Hence we have an immediate algebraic proof of the inequality.