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The algebraic identities (a+b)c \equiv a c + b c and (a+b)^2 \equiv a^2 + 2a b + b^2 can be justified by pictures, as Figures 1 and 2 show.
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 \ge a^2 + b^2 is shown by figure 2.
Figures 4 and 5 demonstrate the inequality a^2 + b^2+ c^2 \ge a b + b c + c a.
In Figure 4, three squares of areas a^2, b^2 and c^2 are shown, assuming (without loss of generality) that a \ge b \ge c.
In Figure 5, three rectangles of areas a b, b c and c a are fitted inside the three squares, showing that a^2 + b^2 + c^2 \ge a b + b c + c a.
Using this inequality, then from Figures 6 and 7 the inequality a^3 + b^3 + c^3 \ge 3a b c can be demonstrated.
The two rectangles have the same base length a + b + c, but the rectangle in Figure 6 has height a^2 + b^2 + c^2, which as we have seen is greater than the height a b + b c + c a 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 (a^2b, b^2c, a^2c, a b^2, b c^2, a c^2). Comparing the remaining areas shows that a^3 + b^3 + c^3 \ge 3a b c, as required.
where x_{1} \ge 0, x_{2} \ge 0, x_{3} \ge 0.
These are special cases of the important Arithmetic Mean - Geometric Mean inequality (the AM-GM inequality)
A re-examination of Figure 3 (see Figure 8) shows that the inequality a^2 + b^2 \ge 2a b becomes an equality if and only if the blue region has zero area.
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 a^2 + b^2 + c^2 \ge a b + b c + c a
is an equality if and only if the two blue regions have zero area.
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 a^3 + b^3 + c^3 \ge 3a b c is an equality if and only if a = b = c.
Proofs with Pictures
The algebraic identities (a+b)c \equiv a c + b c and (a+b)^2 \equiv a^2 + 2a b + b^2 can be justified by pictures, as Figures 1 and 2 show.
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| 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 \ge a^2 + b^2 is shown by figure 2.
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| Figure:3 |
In figure 3, two rectangles, each of area a b, fit inside the two squares of areas a^2 and b^2, showing that a^2 + b^2 \ge 2a b.
In the April 2000 issue of Mathematics Magazine , Claudia Alsina gives further examples of inequalities which can be proved by pictures.
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 a^2 + b^2+ c^2 \ge a b + b c + c a.
In Figure 4, three squares of areas a^2, b^2 and c^2 are shown, assuming (without loss of generality) that a \ge b \ge c.
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| Figure 4 | Figure 5 |
In Figure 5, three rectangles of areas a b, b c and c a are fitted inside the three squares, showing that a^2 + b^2 + c^2 \ge a b + b c + c a.
Using this inequality, then from Figures 6 and 7 the inequality a^3 + b^3 + c^3 \ge 3a b c can be demonstrated.
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| Figure 6 | Figure 7 |
The two rectangles have the same base length a + b + c, but the rectangle in Figure 6 has height a^2 + b^2 + c^2, which as we have seen is greater than the height a b + b c + c a 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 (a^2b, b^2c, a^2c, a b^2, b c^2, a c^2). Comparing the remaining areas shows that a^3 + b^3 + c^3 \ge 3a b c, as required.
The AM-GM inequality
The inequalities a^2 + b^2 \ge 2a b and a^3 + b^3 + c^3 \ge 3a b c can be written in the form\begin{eqnarray} \\ \frac{x_{1} + x_{2}}{2} & \ge & (x_{1}x_{2})^{1/2} \\ \frac{x_{1} + x_{2} + x_{3}}{3} & \ge & (x_{1}x_{2}x_{3})^{1/3} \end{eqnarray}
where x_{1} \ge 0, x_{2} \ge 0, x_{3} \ge 0.
These are special cases of the important Arithmetic Mean - Geometric Mean inequality (the AM-GM inequality)
\begin{eqnarray} \frac{x_{1} + x_{2} + \ldots x_{n}}{n} \ge (x_{1}x_{2}\ldots x_{n})^{(1/n)}, \end{eqnarray}
where
x_{i} \ge 0, 1 \ge i \ge 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 a^2 + b^2 \ge 2a b becomes an equality if and only if the blue region has zero area.
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| 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 a^2 + b^2 + c^2 \ge a b + b c + c a
is an equality if and only if the two blue regions have zero area.
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| 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 a^3 + b^3 + c^3 \ge 3a b c is an equality if and only if a = b = c.
(Remember that throughout this discussion it was assumed that a\ge0, b\ge0 and c\ge0. It can in fact be proved that a^3 + b^3 + c^3 \ge 3a b c under the weaker assumption that a+b+c\ge 0, and that equality holds if and only if a+b+c=0. How? Just look hard at the factorisation a^3+b^3+c^3-3a b c = (a+b+c)(a^2+b^2+c^2-a b-b c-c a)
