Largest possible value of $c$

Question

Let $a$, $b$, $c$ be positive real numbers, with $a + b + 2c = 10$, $c = d$, and $a^2 + b^2 + 2c^2 = 30$. The largest possible value of $c$ can be expressed as $\frac{k + \sqrt{w}}{f}$ in simplified and reduced radical form where $k$, $w$, and $f$ are positive integers. Determine the value of $(k + w + f)$

This is a problem from an old high school math contest(20 questions, 50 minutes), so there is probably some kind of clever algebraic/geometric trick involved. The answer key gives an answer 12, but does not give a solution.

I have substituted $d$ for $c$ and then substituted $c$ into the second equation, though that only gives me an equation in terms of $a$ and $b$. I have also tried completing the square for $a$ and $b$ in the first equation and using that in the second equation, but that always yields extra ab terms.

My intuition tells me I want to minimize $a$ & $b$, but setting them both be 0 does not satisfy the constraints of the two equations. Working algebraically only yields more equations, and since there are 3 variables in two equations, it's difficult to find the maximum of $c$ with non-calculus techniques.

The contest's topic is precalculus, which is why I decided against using Lagrange multipliers(if possible), but I am open to solutions involving more advanced mathematics for my own curiosity's sake.

Answer 1

Addendum added to correct an analytical flaw in my answer.

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Alternative approach.

Constraints (since $c = d$).

• $a + b + (2c) = 10.$
• $a^2 + b^2 + 2(c^2) = 30.$
• $a,b,c > 0.$

See the Addendum, for an explanation of why I am able to infer that the upper bound for $c$ is

$$\frac{5 + \sqrt{5}}{2}. \tag1 $$

The value in (1) above is shown at (4), in the Addendum. It then remains to show that this value is achievable.

That is, I need for

$$\frac{a + b}{2} ~~\text{to equal}~~$$

$$(5 - c) = 5 - \left[\frac{5 + \sqrt{5}}{2}\right] = \frac{5 - \sqrt{5}}{2}. \tag2 $$

I also need for

$$\frac{a^2 + b^2}{2} ~~\text{to equal}~~$$

$$\left(15 - c^2\right) = 15 - \left[\frac{15 + 5\sqrt{5}}{2}\right] = \frac{15 - 5\sqrt{2}}{2}. \tag3 $$

In the pre-Calculus world, you can reason that
when the sum $(a + b)$ is set, and you wish to minimize the sum $(a^2 + b^2)$, you can do so by setting $a = b$.

This may be reasoned by presuming that
$a = k + r, b = k - r \implies (a + b) = 2k,$ some set value. Then
$a^2 + b^2 = k^2 + r^2,$ which is minimized by setting $r = 0$.

Therefore, a reasonable first try at achieving constraints (2) and (3) above is to explore what happens if

$~\displaystyle a = b = \frac{5 - \sqrt{5}}{2},$

which satisfies constraint (2) above.

Then, $~\displaystyle \frac{a^2 + b^2}{2} = a^2 = \frac{25 + 5 - 10\sqrt{5}}{4} = \frac{15 - 5\sqrt{5}}{2},$

which satisfies constraint (3) above.

Therefore, the value of

$$ c = \frac{5 + \sqrt{5}}{2}$$

is achievable, which implies that
$k + w + f = 5 + 5 + 2.$

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Addendum

Originally, I reasoned that :

In general, whatever value is chosen for $c$, this will constrain $a + b = (10 - 2c)$. It is known that given this constraint, the sum $a^2 + b^2$ is minimized, when $a = b$. Minimizing $(a^2 + b^2)$ is the actual goal, since this allows $c^2$ to be maximized.

An easy way to see this is in the pre-Calculus world is to consider $a = (k + r), b = (k - r) \implies (a + b) = 2k$, which is presumed to be some set value.
Then $a^2 + b^2 = 2k^2 + 2r^2$, which is minimized when $r = 0.$

Thanks to dxiv for commenting that the above analysis is invalid. That is, based on the above analysis, you could have that

• $\displaystyle c > \frac{5 + \sqrt{5}}{2}$,

• $(a + b) < \left(5 - \sqrt{5}\right),$

• and $a \neq b.$

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$\underline{\text{Corrected Analysis}}:$

I will use the comment of dxiv (which follows my answer) as a guide:

For any $0 < a,b$, you have that

$a^2 - 2ab + b^2 = (a - b)^2 \geq 0 \implies a^2 + b^2 \geq 2ab.$

Therefore,

$\displaystyle \frac{a^2 + b^2}{2} \geq \frac{a^2 + 2ab + b^2}{4} \implies \sqrt{\frac{a^2 + b^2}{2}} \geq \frac{a + b}{2}.$

Since the constraints imply that

• $\displaystyle \frac{a + b}{2} = (5 - c)$

• and $~\displaystyle \frac{a^2 + b^2}{2} = (15 - c^2) \implies $
  $~\displaystyle \sqrt{15 - c^2} = \sqrt{\frac{a^2 + b^2}{2}} \geq \frac{a + b}{2} = (5 - c)$

You have that

$$(15 - c^2) \geq (5 - c)^2 = 25 - 10c + c^2 \implies $$

$$2c^2 - 10c + 10 \leq 0 \implies c^2 - 5c + 5 \leq 0.$$

This implies that

$$\left(c - \frac{5}{2}\right)^2 - \frac{25}{4} + 5 \leq 0.$$

This implies that

$$\left(c - \frac{5}{2}\right)^2 - \frac{5}{4} \leq 0.$$

This implies that

$$\left(c - \frac{5}{2}\right) \leq \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.$$

This implies that $c$ can be no bigger than

$$\frac{5}{2} + \frac{\sqrt{5}}{2}. \tag4 $$

Answer 2

Given $c,$ you get $$a+b=10-2c,\\a^2+b^2=30-2c^2$$

From this, you get $$\begin{align} ab&=\frac{(a+b)^2-(a^2+b^2)}2\\&=\frac{(10-2c)^2-(30-2c^2)}{2}\\&=35-20c+3c^2 \end{align} $$

Solving $x^2-(a+b)x+ab=0$ gives $$\begin{align} a,b &=\frac{(10-2c)\pm \sqrt{(10-2c)^2-4(35-20c+3c^2)}}2\\ &=(5-c)\pm \sqrt{10c-10-2c^2} \end{align} $$

So you need $$c^2-5c+5\leq 0,\tag1$$ to get $a,b$ real, and you need $$5-c\geq \sqrt{10c-10-2c^2}\tag2$$ to ensure $a,b$ are positive.

$(1)$ gives $$\frac{5-\sqrt{5}}{2}\leq c\leq \frac{5+\sqrt{5}}{2}.$$

When $c=\frac{5+\sqrt{5}}2,$ you get $a=b=\frac{5-\sqrt5}2.$

Answer 3

The other answers are brilliant, but for completeness I'm adding an answer based on dxiv's hint, because it seems to yield the fastest solution. For any $a,b$ we have $(a+b)^2\leq 2(a^2+b^2)$, because $$(a+b)^2 = a^2+b^2+2ab\leq a^2+b^2+(a^2+b^2)=2(a^2+b^2).$$

Therefore, from $a+b+2c=10$ and $a^2+b^2+2c^2=30$ we find $$(10-2c)^2\leq 2(30-2c^2)$$ and solving this quadratic inequality gives $$\frac{5-\sqrt{5}}{2}\leq c\leq \frac{5+\sqrt{5}}{2}.$$

This means that every solution $(a,b,c)$ must satisfy $c\leq \frac{5+\sqrt{5}}{2}$. To see that $c=\frac{5+\sqrt{5}}{2}$ is an actual solution, we can find $a,b$ by solving $$\begin{cases}a+b=10-2c=10-(5+\sqrt{5})=5-\sqrt{5}\\ a^2+b^2=30-2c^2=30-2\left(\frac{5+\sqrt{5}}{2}\right)^2=15-5\sqrt{5} \end{cases}$$ so substituting $b=5-\sqrt{5}-a$ into the second equation we get the quadratic equation $a^2+(5-\sqrt{5}-a)^2=15-5\sqrt{5}$ which has the unique solution $a=\frac{5-\sqrt{5}}{2}$ and by symmetry (or by substitution) we get also $b=\frac{5-\sqrt{5}}{2}$. Since $a,b$ are indeed positive real numbers this verifies that $c=\frac{5+\sqrt{5}}{2}$ is a solution and so it is the maximal value.

Answer 4

$\underline{\text{Overview}}$

In this answer, I intentionally limit my analysis to pre-Calculus methods.
This (overhauled) answer dissects the following problem.

Given the following:

• $S,T$ are fixed positive constants.
• $a,b,c$ are all variables under the following constraints:
• $a,b,c > 0$.
• $a + b + 2c = S$.
• $a^2 + b^2 + 2c^2 = T$.
• Without loss of generality, assume that $a \geq b.$

The goal is to identify the least upper bound for $c$ that is attainable, where there is an ordered triple $(a,b,c)$ that satisfies the constraints.

This answer will explore the relationship between $S$ and $T$ to answer the following questions:

• Under what circumstances will it be impossible for any $(a,b,c)$ to satisfy the constraints?

• Under what circumstances will it be possible for some $(a,b,c)$ to satisfy the constraints, but where the least upper bound for $c$ can only be approached, but never obtained? In these circumstances, what will the form of a satisfying $(a,b,c)$ take, where $c$ approaches the least upper bound?

• Under what circumstances will it be possible for $(a,b,c)$ to satisfy the constraints, where the least upper bound for $c$ can be obtained? In these circumstances, what form will a satisfying $(a,b,c)$ take?

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Lemma 1
Given the following constraints:

• $M$ is a fixed positive number, $~x,y~$ are variables.
• $0 < y \leq x.$
• $x + y = M.$

Then,

• The minimum value for $x^2 + y^2$ is $~\displaystyle \frac{M^2}{2},~$ which is obtainable only by setting $~\displaystyle x = y = \frac{M}{2}.$

• The least upper bound for $(x^2 + y^2)$ is $~M^2,~$ which may be approached (but never attained) by having $y$ approach $0$, and $x$ approach $M$.

$\underline{\text{Proof}}$
Let $~\displaystyle r = x - \frac{M}{2} \implies ~0 \leq r < \frac{M}{2}, ~x = \left(\frac{M}{2} + r\right), ~y = \left(\frac{M}{2} - r\right).$

Then $~\displaystyle (x^2 + y^2) = \frac{M^2}{2} + 2r^2.$
Clearly, the sum $(x^2 + y^2)$ is minimized by setting $r = 0.$
This can only occur when $~\displaystyle x = y = \frac{M}{2}.$

Since the (unobtainable) least upper bound for $r$ is $~\displaystyle \frac{M}{2}$,
the (unobtainable) least upper bound for the sum $(x^2 + y^2)$ is $M^2$,
which is approached when $y$ approaches $0$ and $x$ approaches $M$.

Lemma 2
Given the following constraints:

• $M$ is a fixed positive number, $~x,y~$ are variables.
• $0 < x,y.$
• $x + 2y = M.$

Then, the least upper bound for $x^2 + 2y^2$ is $~M^2,~$ which may be approached (but never attained) by having $y$ approach $0$, and $x$ approach $M$.

$\underline{\text{Proof}}$
For any given value of $x$, where $0 < x < M,$ you have that

$3x^2 - 2Mx < 3x^2 - 2x^2 = x^2 < M^2 \implies $

$~\displaystyle \frac{3x^2}{2} - Mx < \frac{M^2}{2} \implies $

$~\displaystyle \frac{3x^2}{2} - Mx + \frac{M^2}{2} < M^2 \implies $

$~\displaystyle x^2 + 2\left(\frac{M^2 - 2Mx + x^2}{4}\right) < M^2 \implies $

$~\displaystyle x^2 + 2 \left[\frac{M - x}{2}\right]^2 < M^2 \implies $

$~\displaystyle x^2 + 2y^2 < M^2.$

Clearly, as $x$ approaches $M$ (from below) and $y$ approaches $0$, then $(x^2 + 2y^2)$ is approaching $~M^2,~$ from below.

Therefore, the (unobtainable) least upper bound of $x^2 + 2y^2 = M^2$ can only be approached, by having $x$ approach $M$ and $y$ approach $0$.

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$\underline{\text{Minimum and Maximum Values for }~ T}$

Given that $~0 < a,b,c,~$ and $~a + b + 2c = S$,
what are the minimum and maximum possible values for $T = a^2 + b^2 + 2c^2?$

Let $~\displaystyle k = \frac{a + b}{2} \implies (a + b) = 2k.$

Under the constraint $a + b + 2c = S$:

• It is then implied that $~\displaystyle c = \left(\frac{S}{2} - k\right).$
• Let $f(k)$ denote the greatest lower bound for $T = a^2 + b^2 + 2c^2$, as a function of $k$.
• Let $g(k)$ denote the least upper bound for $T = a^2 + b^2 + 2c^2$, as a function of $k$.

Note that for any (fixed) value of $k$, the value of $c$ is unaffected by which values are chosen for $a,b$ so that $a + b = 2k$.

To Minimize $T$:

By Lemma 1, $~\displaystyle f(k) = \frac{(2k)^2}{2} + 2c^2 = 2k^2 + 2\left(\frac{S}{2} - k\right)^2,~$ which is attained when $a = b = k$.

So, when determining the value of $k$ that results in the minimum value of $T$, it should be assumed that $a = b$.

This simplifies the problem to finding the appropriate choices for $a,c$ so that

$~\displaystyle a + c = \frac{S}{2},~$ and the sum $(a^2 + c^2)$ is minimized.

Again invoking Lemma 1, $~\displaystyle \frac{T}{2} = a^2 + c^2~$ is minimized when $~\displaystyle a = c = \frac{S}{4}.$

This implies that the minimum value for $T = a^2 + b^2 + 2c^2$ is

$\displaystyle 4 \times \frac{S^2}{16} = \frac{S^2}{4},~$ which is achieved when

$~\displaystyle a = b = c = \frac{S}{4}.$

To Maximize $T$:

By Lemma 1, $~\displaystyle g(k) = (2k)^2 + 2c^2 = 2k^2 + \left(\frac{S}{2} - k\right)^2,~$ which is approached (but never attained) when $a$ approaches $2k$ and $b$ approaches $0$.

So, when determining the value of $k$ that results in the (unobtainable) least upper bound for $T$, it should be assumed that $a = 2k$, and $b = 0$.

This simplifies the problem to finding the appropriate choices for $a,c$ so that $a + 2c = S,~$ and $T = a^2 + 2c^2$ is maximized.

By Lemma 2, the (unobtainable) least upper bound of $T = S^2$ is approached by having $a$ approach $S$ and $c$ approach $0$.

In Summary

If $~\displaystyle T < \frac{S^2}{4}~$ or $~\displaystyle T \geq S^2$,
then the constraints can not be satisfied.

Therefore, without loss of generality, the remainder of this problem will assume that

$$\frac{S^2}{4} \leq T < S^2. \tag1 $$

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$\underline{\text{Upper and Lower Bounds for }~ c}$

This section highjacks and generalizes the analysis in the answer given by Thomas Andrews. I would never have thought of this analysis on my own.

$a + b = S - 2c$.
$a^2 + b^2 = T - 2c^2.$

Therefore, $\displaystyle ab = \frac{(a + b)^2 - (a^2 + b^2)}{2} = \frac{(S - 2c)^2 - (T - 2c^2)}{2}.$

Solving for $~a,b~$ by $x^2 - (a + b)x + ab = 0$ gives

$\displaystyle x = \frac{1}{2} \left[(a + b) \pm \sqrt{(a + b)^2 - 4ab} \right]$

$\displaystyle = \frac{1}{2} \left[(S - 2c) \pm \sqrt{(S - 2c)^2 - 4\left(\frac{1}{2}\right)\left[(S - 2c)^2 - (T - 2c^2)\right]} \right]$

$\displaystyle = \frac{1}{2} \left[(S - 2c) \pm \sqrt{2(T - 2c^2) - (S - 2c)^2}\right]$

$\displaystyle = \frac{1}{2} \left[(S - 2c) \pm \sqrt{(2T - 4c^2) - (S^2 - 4Sc + 4c^2)}\right]$

$\displaystyle = \frac{1}{2} \left[(S - 2c) \pm \sqrt{(2T - S^2) + 4Sc - 8c^2}.\right]$

This implies that

$$a,b = \left(\frac{S}{2} - c\right) \pm \sqrt{-2c^2 + Sc + \left(\frac{T}{2} - \frac{S^2}{4}\right)}. \tag2 $$

So, in order for $a,b$ to both be positive, you need that

$$-2c^2 + Sc + \left(\frac{T}{2} - \frac{S^2}{4}\right) \geq 0. \tag3 $$

and

$$\left(\frac{S}{2} - c\right) - \left[-2c^2 + Sc + \left(\frac{T}{2} - \frac{S^2}{4}\right)\right] > 0. \tag4 $$

Exploring (3) above, you need that

$\displaystyle 4c^2 - 2Sc + \left(\frac{S^2}{2} - T\right) \leq 0 \iff $

$\displaystyle \left[ \left(2c - \frac{S}{2}\right)^2 - \frac{S^2}{4}\right] + \left(\frac{S^2}{2} - T\right) \leq 0 \iff $

$\displaystyle \left(2c - \frac{S}{2}\right)^2 + \left(\frac{S^2}{4} - T\right) \leq 0 \iff $

$$- \sqrt{T - \frac{S^2}{4}} \leq \left(2c - \frac{S}{2}\right) \leq \sqrt{T - \frac{S^2}{4}}. \tag5 $$

Using (1) above, from the previous section, the analysis will now be split into two cases, depending on whether

• $\displaystyle \frac{S^2}{4} \leq T < \frac{S^2}{2}.$

• or $\displaystyle \frac{S^2}{2} \leq T < S^2.$

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$\underline{\text{Case 1:} ~~\displaystyle \frac{S^2}{4} \leq T < \frac{S^2}{2}}$

Note that

$\displaystyle \frac{S^2}{2} > T \geq \frac{S^2}{4} \implies $

$\displaystyle \frac{S^2}{4} > T - \frac{S^2}{4} \geq 0 \implies $

$$\frac{S}{2} > \sqrt{T - \frac{S^2}{4}}. \tag6 $$

From (5) above, in the "Upper and Lower Bounds..." section, the least upper bound for $c$ is

$$ c = \left[\frac{1}{2}\right] \times \left[\frac{S}{2} + \sqrt{T - \frac{S^2}{4}}\right]. \tag7 $$

Note that by (6) above, $~c~$ is less than

$~\displaystyle \left[\frac{1}{2}\right] \times \left[2 \times \frac{S}{2}\right] = \frac{S}{2}.$

This implies that $~\displaystyle \left[\frac{S}{2} - c\right] > 0.$

With $c$ held at this fixed positive value shown in (7) above, the sum $(a + b)$ is fixed at the value of $(S - 2c)$. By Lemma 1, the sum of $(a^2 + b^2)$ may then be minimized by setting both $a$ and $b$ to the positive value of

$\displaystyle \left(\frac{S}{2} - c\right) = \left[\frac{1}{2}\right] \times \left[\frac{S}{2} - \sqrt{T - \frac{S^2}{4}}\right].$

This implies that

$$(a^2 + b^2) = \left(\frac{1}{2}\right) \times \left\{ ~\frac{S^2}{4} - S\sqrt{T - \frac{S^2}{4}} + \left(T - \frac{S^2}{4}\right) ~\right\}.$$

Therefore, $(a^2 + b^2)$ equals

$$\left(\frac{1}{2}\right) \times \left[T - S\sqrt{T - \frac{S^2}{4}}\right].$$

Similarly, $2c^2$ equals

$$\left(\frac{1}{2}\right) \times \left[T + S\sqrt{T - \frac{S^2}{4}}\right].$$

Therefore $a^2 + b^2 + 2c^2 = T,$ as required.

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$\underline{\text{Case 2:}~~ \displaystyle \frac{S^2}{2} \leq T < S^2}$

This case is very tricky. As will be shown in this section, satisfying values for $a,b,c$ can always be found for any $T$ in this range. Also, the least upper bound for $c$ will be a function of the particular value of $T$ in this range. However, for all $T$ in this range, the least upper bound for $c$ can be approached, but never attained.

Since $~\displaystyle T \geq \frac{S^2}{2}~$ the value of $~\displaystyle \left(\frac{S}{2} - c\right)~$ is no longer positive. Therefore, you can't have $~(a + b) = (S - 2c),~$ based on the value of $c$ shown in (7). Therefore, a totally different approach is needed here.

Suppose that $~\displaystyle c = \frac{S}{2} - n ~: 0 < n < \frac{S}{2}.$

Then $~\displaystyle 2c^2 = \left[\frac{S^2}{2} - 2Sn + 2n^2\right].$

There are now two constraints on $a,b$:

• $(a + b) = S - 2c = 2n$.

• $\displaystyle (a^2 + b^2) = T - 2c^2 = \left(T - \frac{S^2}{2}\right) + 2Sn - 2n^2.$

By Lemma 1, the range of all possible values for the sum $(a^2 + b^2)$ is
$2n^2 \leq (a^2 + b^2) < 4n^2.$

To maximize $c$, $n$ must be chosen as small as possible so that the following constraint is achieved:

$$2n^2 \leq \left(T - \frac{S^2}{2}\right) + 2Sn - 2n^2 < 4n^2.$$

This implies that

$$4n^2 - 2Sn - \left(T - \frac{S^2}{2}\right) \leq 0 < 6n^2 - 2Sn - \left(T - \frac{S^2}{2}\right). \tag8 $$

First, consider the LHS constraint in (8) above.

It can be re-written as :

$\displaystyle \left[ ~\left(2n - \frac{S}{2}\right)^2 - \frac{S^2}{4} ~\right] - T + \frac{S^2}{2} \leq 0 \iff $

$$\left(2n - \frac{S}{2}\right)^2 \leq T - \frac{S^2}{4}. \tag{9}$$

However, because $~\displaystyle 0 < n < \frac{S}{2}, ~$ the LHS of the constraint in (9) above

must be less than $\displaystyle \frac{S^2}{4}.$

Also, since in this section, $T$ is assumed $~\displaystyle \geq \frac{S^2}{2},~$ the RHS of (9) above

must be $~\displaystyle \geq \frac{S^2}{4}.$

Therefore, the LHS constraint in (8) will be automatically satisfied for any

$n$ such that $~\displaystyle 0 < n < \frac{S}{2}.$

The RHS constraint in (8) may be re-written as

$\displaystyle 0 < 36n^2 - 12Sn - 6T + 3S^2.$

This may be re-expressed as

$$0 < [(6n - S)^2 - S^2] - 6T + 3S^2 \iff 6T - 2S^2 < (6n - S)^2. \tag{10} $$

This means that either

$$6n - S < - \sqrt{6T - 2S^2} \implies 6n < S - \sqrt{6T - 2S^2} \tag{11} $$

or

$$6n - S > \sqrt{6T - 2S^2} \implies 6n > S + \sqrt{6T - 2S^2} \tag{12}.$$

However, since $~\displaystyle \frac{S^2}{2} \leq T < S^2,$ you have that

$\displaystyle S^2 \leq (6T - 2S^2) < 4S^2.$

This implies that $S \leq \sqrt{6T - 2S^2} < 2S.$

This implies that $~\displaystyle -S < S - \sqrt{6T - 2S^2} \leq 0.$

Therefore, since $6n > 0,~$ it is impossible for (11) above to be satisfied.

Therefore, $n$ will have to be chosen so that (12) above is satisfied.

This implies that for a specific value of $T$, the optimal (minimum) value for $n$, which can be approached (from above) but not attained, is

$$n = \frac{S}{6} + \frac{\sqrt{6T - 2S^2}}{6}. \tag{13} $$

Based on (13) above, the (unattainable) least upper bound for $c$ is

$$\frac{S}{2} - n = \frac{S}{3} - \frac{\sqrt{6T - 2S^2}}{6}. \tag{14}$$

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$\underline{\text{Case 2: Shortcut and Wrap Up}}$

If you review the previous section, you will see that the choice of the minimum value of $n$ to satisfy (12) above is based on the idea that

With $(a + b) = 2n,~$ $~(a^2 + b^2) ~$ could approach the value $4n^2$ from below.

This means that $a$ could approach the value $2n$ from below, as $b$ approached $0$.

This implies that for any $T$ in the pertinent Case 2 range, the unachievable value for $n$ will have form

$\displaystyle a = 2n, ~c = \frac{s}{2} - n.$

Using (13) and (14) from the previous section, this gives the unobtainable

• $\displaystyle a = \frac{S}{3} + \frac{\sqrt{6T - 2S^2}}{3}.$

• $\displaystyle c = \frac{S}{3} - \frac{\sqrt{6T - 2S^2}}{6}.$

You can tell at a glance that $a + 2c = S$, and the equation

$a^2 + 2c^2 = T$ is easily verifiable.

However, the real point of this section is that it is arguable whether the analysis of the previous section was necessary. Some method had to be found to demonstrate that the Least Upper Bound on $c$ would have to be based on $a = 2n, b = 0$.

Once that is done, the values for $a,c$ can more easily be derived. However, I couldn't think of any alternative, clear cut proof, besides the (long winded) previous section.