A curious consequence of the implication (iv) \(\Rightarrow\) (iii) of (4.11) is the following numerical fact: Given integers \(a, b_1, \ldots, b_6\) such that \(b_i>0\) for each \(i\), \(a-b_i-b_j>0\) for each \(i, j\) and \(2 a-\sum_{i \neq j} b_i>0\) for each \(j\), we must necessarily have \(a^2-\sum b_i^2>0\). Prove this directly (for \(a, b_1, \ldots, b_6 \in \RR\)) using methods of freshman calculus.
Solution
We use only the inequalities in the statement. The pairwise condition is needed only for distinct indices; if the phrase ``for each \(i,j\)’’ is read as also including \(i=j\), that is a stronger hypothesis.
1. We first prove the following elementary lemma. Let\[ A>0, \]and let\[ x_1,\ldots,x_6>0 \]satisfy\[ x_i+x_j<A \qquad(i\ne j) \]and\[ \sum_{i=1}^6x_i\le2A. \]Then\[ \boxed{\sum_{i=1}^6x_i^2<A^2.} \]
Reorder the variables so that \[ x_1\ge x_2\ge\cdots\ge x_6>0, \] and put \[ S=\sum_i x_i. \]
First suppose \[ x_1\le\frac A2. \] Then \[ \sum_i x_i^2 \le x_1\sum_i x_i \le \frac A2\,2A =A^2. \] The inequality is strict: equality would require both \(S=2A\) and \(x_i=x_1=A/2\) for every \(i\) with \(x_i>0\), which would contradict \[ x_1+x_2<A. \]
Now suppose \[ x_1>\frac A2. \] Since \(x_i\le x_2\) for \(i\ge2\), \[ \sum_i x_i^2 \le x_1^2+x_2(S-x_1). \] The hypotheses give \[ x_2<A-x_1 \] and \[ S-x_1\le2A-x_1. \] Hence \[ \sum_i x_i^2 < x_1^2+(A-x_1)(2A-x_1). \] But \[ \begin{aligned} &x_1^2+(A-x_1)(2A-x_1)-A^2\\ &\qquad= (2x_1-A)(x_1-A). \end{aligned} \] Here \[ 2x_1-A>0, \] while \(x_1<A\) because \(x_2>0\) and \(x_1+x_2<A\). Thus \[ (2x_1-A)(x_1-A)<0, \] and again \[ \sum_i x_i^2<A^2. \] This proves the lemma.
2. Put\[ s=\sum_{i=1}^6 b_i. \]The hypotheses imply\[ \boxed{a>0.} \]
Every \(b_i\) is positive, and for two distinct indices \(i,j\) one has \[ a>b_i+b_j>0. \]
3. If\[ s\le2a, \]then\[ \boxed{\sum_i b_i^2<a^2.} \]
Apply the lemma of step 1 with \[ A=a, \qquad x_i=b_i. \] The hypotheses \[ b_i>0, \qquad b_i+b_j<a \] are exactly the positivity and pairwise hypotheses of the lemma, while the present case assumption is \[ \sum_i b_i=s\le2a. \] Therefore \[ \sum_i b_i^2<a^2. \]
4. It remains to treat the case\[ s>2a. \]Set\[ t=s-2a>0 \]and\[ c_i=b_i-t. \]Then\[ \boxed{c_i>0\text{ for every }i.} \]
For each \(i\), the third family of hypotheses gives \[ 2a-\sum_{j\ne i}b_j>0. \] Since \[ \sum_{j\ne i}b_j=s-b_i, \] this says \[ b_i>s-2a=t. \] Hence \[ c_i=b_i-t>0. \]
5. Put\[ A=a-2t. \]Then\[ \boxed{A>0} \]and, for distinct \(i,j\),\[ \boxed{c_i+c_j<A.} \]
Using the pairwise hypothesis, \[ \begin{aligned} c_i+c_j &=b_i+b_j-2t\\ &<a-2t\\ &=A. \end{aligned} \] The left side is positive by step 4, so necessarily \(A>0\).
6. The translated variables also satisfy\[ \boxed{\sum_i c_i<2A.} \]
Since \[ s=2a+t, \] one has \[ \sum_i c_i = s-6t = 2a-5t. \] On the other hand, \[ 2A=2a-4t. \] Because \(t>0\), \[ 2a-5t<2a-4t=2A. \]
7. Therefore\[ \boxed{\sum_i c_i^2<A^2.} \]
Steps <1>4–<1>6 verify all hypotheses of the lemma in step 1 for the six numbers \(c_i\) and the positive number \(A=a-2t\). Hence \[ \sum_i c_i^2<A^2. \]
8. In the case \(s>2a\) one still has\[ \boxed{\sum_i b_i^2<a^2.} \]
Because \[ b_i=c_i+t, \] step 7 gives \[ \begin{aligned} \sum_i b_i^2 &= \sum_i(c_i+t)^2\\ &= \sum_i c_i^2 +2t\sum_i c_i +6t^2\\ &< A^2 +2t(2a-5t) +6t^2. \end{aligned} \] Now substitute \[ A=a-2t. \] Then \[ \begin{aligned} A^2+2t(2a-5t)+6t^2 &=(a-2t)^2+4at-10t^2+6t^2\\ &=a^2. \end{aligned} \] Thus \[ \sum_i b_i^2<a^2. \]
9. Consequently\[ \boxed{a^2-\sum_{i=1}^6b_i^2>0.} \]
3. If \(s>2a\), it is step
8. These two casesexhaust all possibilities.
10. Q.E.D.
Step 1 proves the elementary auxiliary inequality, and steps <1>2–<1>9 apply it directly to the hypotheses of the problem.