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Solution to puzzle 114: Sums of squares and cubes

Let a, b, and c be positive real numbers such that abc = 1.  Show that a2 + b2 + c2 less than or equal to a3 + b3 + c3.


The Rearrangement Inequality

The rearrangement inequality, stated without proof below, is intuitively quite clear.

Let a1 less than or equal to a2 less than or equal to ... less than or equal to an and b1 less than or equal to b2 less than or equal to ... less than or equal to bn be real numbers.  For any permutation (r1, r2, ..., rn) of (b1, b2, ..., bn), we have:

a1b1 + a2b2 + ... + anbn  greater than or equal to  a1r1 + a2r2 + ... + anrn  greater than or equal to  a1bn + a2bn−1 + ... + anb1,

with equality if, and only if, (r1, r2, ..., rn) is equal to (b1, b2, ..., bn) or (bn, bn−1, ..., b1), respectively.

That is, the sum is maximal when the two sequences, {ai} and {bi}, are sorted in the same way, and is minimal when they are sorted oppositely.

We will use an extension of the rearrangement inequality to three sequences.  This extension has three qualifications:

So the statement of the extension is as given below.  Although the extension may be regarded as a standard result, a proof is also given.

Let 0 less than or equal to a1 less than or equal to a2 less than or equal to ... less than or equal to an0 less than or equal to b1 less than or equal to b2 less than or equal to ... less than or equal to bn, and 0 less than or equal to c1 less than or equal to c2 less than or equal to ... less than or equal to cn be real numbers.
For any permutations (r1, r2, ..., rn) of (b1, b2, ..., bn), and (s1, s2, ..., sn) of (c1, c2, ..., cn), we have:

a1b1c1 + a2b2c2 + ... + anbncn  greater than or equal to  a1r1s1 + a2r2s2 + ... + anrnsn.

That is, the sum is maximal when the three sequences, {ai}, {bi}, and {ci} are sorted in the same way.

Proof

The proof will be by mathematical induction on n.  The result is trivially true for n = 1.

For the induction step, we show that if the result is true for some arbitrary value, n = k − 1, then it also holds for n = k.  (Where k > 0.)

Suppose the largest terms in each sequence, ak, bk, and ck are not together in the same product.
Suppose also that ak and bk are found in the terms akbpcq + arbkcs.(1)
Suppose, without loss of generality, that cq less than or equal to cs.  Then the terms of the b sequence and the c sequence are increasing, but those of the a sequence are decreasing.
If we write (1) as ak(bpcq) + ar(bkcs), and consider the bracketed terms as single terms, then, since all terms are non-negative, we have bpcq less than or equal to bkcs, and we can apply the rearrangement inequality for two sequences.
Hence arbpcq + akbkcs greater than or equal to akbpcq + arbkcs.

Using this technique, we may bring together (if they are apart) the terms ak, bk, and ck in one product, without diminishing the sum.  The technique also shows that no larger sum may be formed with ak, bk, and ck not together in the same product.  Hence the largest sum must occur when ak, bk, and ck are together.  (Since the sorting process is finite, it converges, and the maximum is actually found.)

Hence, by the inductive hypothesis, the result is true for k.  This concludes the proof by induction.

abc = 1 implies a2 + b2 + c2 less than or equal to a3 + b3 + c3

Now consider a2 + b2 + c2 less than or equal to a3 + b3 + c3.
We homogenize the inequality by incorporating the side condition, that is, by multiplying the left-hand side by 1 = (abc)1/3, yielding the candidate inequality below, in which all terms are now of degree 3.
a7/3b1/3c1/3 + a1/3b7/3c1/3 + a1/3b1/3c7/3 less than or equal to a3 + b3 + c3.(1)
If we can prove this inequality, then the target inequality will follow upon division by 1 = (abc)1/3.

Assume without loss of generality that a less than or equal to b less than or equal to c.
Then the sequences
{a7/3, b7/3, c7/3},
{a1/3, b1/3, c1/3},
{a1/3, b1/3, c1/3},
are sorted in the same way, while the sequences
{a7/3, b7/3, c7/3},
{b1/3, c1/3, a1/3},
{c1/3, a1/3, b1/3},
are not sorted the same way.

Applying the rearrangement inequality for three sequences to the above sequences, we obtain (1).
We then divide the left-hand side of (1) by 1 = (abc)1/3, obtaining
a2 + b2 + c2 less than or equal to a3 + b3 + c3.

Therefore, if abc = 1, we conclude a2 + b2 + c2 less than or equal to a3 + b3 + c3.


Remarks

Clearly the above proof may be generalized to show that, if abc = 1, then, for any positive integer n:
3 less than or equal to a + b + c less than or equal to a2 + b2 + c2 less than or equal to a3 + b3 + c3 less than or equal to a4 + b4 + c4 less than or equal to ... less than or equal to an + bn + cn.

Inequality (1) may also be proved using Muirhead's Inequality.


A simpler solution

Greg Muller sent the following ingenious solution.

Lemma

For R a positive real number and n a positive integer, Rn(R − 1) greater than or equal to (R − 1), with equality if and only if R = 1.

Proof

Consider the three cases, R > 1, R = 1, R < 1, all of which are straightforward.

We must now show that abc = 1 implies a2 + b2 + c2 less than or equal to a3 + b3 + c3.
This is equivalent to showing that a3 + b3 + c3 − a2 − b2 − c2 greater than or equal to 0.

But a3 + b3 + c3 − a2 − b2 − c2 = a2(a − 1) + b2(b − 1) + c2(c − 1) greater than or equal to (a − 1) + (b − 1) + (c − 1) = (a + b + c) − 3.

By the Arithmetic Mean-Geometric Mean Inequality, (a + b + c)/3 greater than or equal to (abc)1/3 = 1.
Hence a + b + c greater than or equal to 3.

Therefore, a3 + b3 + c3 − a2 − b2 − c2 greater than or equal to 0, and the result follows.


Additional problem

Let a, b, and c be positive real numbers such that a + b + c = 1.  Show that a2 + b2 + c2 less than or equal to 3(a3 + b3 + c3).


Further reading

  1. (Adobe) Portable Document Format The Rearrangement Inequality by K. Wu and Andy Liu -- a tutorial that shows how to derive many other inequalities, such as Arithmetic Mean - Geometric Mean, Geometric Mean - Harmonic Mean, and Cauchy-Schwartz, from the Rearrangement Inequality

Source: The Cauchy-Schwarz Master Class, by J. Michael Steele. See Chapter 12.  With special thanks to Andy Liu for help in proving the rearrangement inequality for three sequences.

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