Proof with pictures

                             Proof with pictures

The algebraic identities (a+b)c=ac+bc and (a+b)2=a2+2ab+b2 can be justified by pictures, as Figures 1 and 2 show. 

Figure 1		Figure 2

Arguments of this nature can be found in Euclid's The Elements (book II) . Like Euclid, we will assume throughout this discussion that a, b and c are non-negative. 

Inequalities can also be demonstrated by pictures. For example, the inequality (a+b)2≥a2+b2is shown by figure 2. 

Figure:3

In figure 3, two rectangles, each of area ab, fit inside the two squares of areas a2 and b2, showing that a2+b2≥2ab.

In the April 2000 issue of Mathematics Magazine , Claudia Alsina gives further examples of inequalities which can be proved by pictures.

Figures 4 and 5 demonstrate the inequality a2+b2+c2≥ab+bc+ca. 

In Figure 4, three squares of areas a2, b2 and c2 are shown, assuming (without loss of generality) that a≥b≥c.

Figure 4		Figure 5

In Figure 5, three rectangles of areas ab, bc and ca are fitted inside the three squares, showing that a2+b2+c2≥ab+bc+ca. 

Using this inequality, then from Figures 6 and 7 the inequality a3+b3+c3≥3abc can be demonstrated.

Figure 6		Figure 7

The two rectangles have the same base length a+b+c, but the rectangle in Figure 6 has height a2+b2+c2, which as we have seen is greater than the height ab+bc+ca of the rectangle in Figure 7. So the area of the rectangle in Figure 6 is greater than the area of the rectangle in Figure 7. 

The two rectangles are each divided into nine small rectangles, with areas as shown. The six green rectangles in Figure 6 have the same areas as the six green rectangles in Figure 7 (a2b, b2c, a2c, ab2, bc2, ac2). Comparing the remaining areas shows that a3+b3+c3≥3abc, as required.

The AM-GM inequality

The inequalities a2+b2≥2ab and a3+b3+c3≥3abc can be written in the form

x1+x22x1+x2+x33≥≥(x1x2)1/2(x1x2x3)1/3

where

x1≥0,x2≥0,x3≥0.

These are special cases of the important Arithmetic Mean - Geometric Mean inequality (the AM-GM inequality)

x1+x2+…xnn≥(x1x2…xn)(1/n),

where

xi≥0,1≥i≥n

When is an inequality an equality?

When an inequality is established, it is always important to know under what circumstances equality can occur.

A re-examination of Figure 3 (see Figure 8) shows that the inequality a2+b2≥2ab becomes an equality if and only if the blue region has zero area. 

Figure: 8

The blue region is a square of area (a−b)2, which is zero if and only if a=b.

Similarly, a re-examination of Figure 5 (see Figure 9) shows that the inequality

a2+b2+c2≥ab+bc+ca

is an equality if and only if the two blue regions have zero area.

Figure: 9

This occurs if and only if (a−b)2=(a−c)(b−c)=0,

i.e. if a=b=c

It follows that the inequality a3+b3+c3≥3abc is an equality if and only if a=b=c.

(Remember that throughout this discussion it was assumed that a≥0, b≥0 and c≥0. It can in fact be proved that a3+b3+c3≥3abc under the weaker assumption that a+b+c≥0, and that equality holds if and only if a+b+c=0. How? Just look hard at the factorisation

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)