Inspired by Sharky Kesa

If $a+b+c=k > 0$ and $ab + ac + bc = \tfrac14k^2$ , then $a,b,c$ are the real roots of the cubic equation $f(X) \; = \; X^3 - kX^2 + \tfrac14k^2X - q \; = \; 0$ for some real $q$ . Now $f'(X) \; = \; 3X^2 - 2kX + \tfrac14k^2 \; = \; \big(X - \tfrac12k\big)\big(3X - \tfrac12k\big)$ and so $f$ has turning points at $\tfrac16k$ and $\tfrac12k$ . Since $f\big(\tfrac16k\big) = \tfrac{k^3}{54} - q$ and $f\big(\tfrac12k\big) = -q$ , we deduce that $0 \le q \le \tfrac{1}{54}k^3$ .

When $q=0$ , the roots of $f(X) = 0$ are $\tfrac12k,\tfrac12k,0$ and so the least possible value of $c$ is $0$ .

When $q = \tfrac{1}{54}k^3$ , the roots of $f(X)= 0$ are $\tfrac16k,\tfrac16k, \tfrac23k$ , and so the largest possible value of $c$ is $\tfrac23k$ . This makes the answer in this case equal to $\boxed{82.304}$ .

If $k < 0$ , then the largest value of $c$ is $0$ and the least value of $c$ is $\tfrac23k$ , and so the sum of the smallest and largest values of $c$ is still $\tfrac23k$ .

solving for c $c=123.456-a-b\to ab+(123.456-a-b)(a+b)=3810.345984$ we differentiate the second equation w.r.t b and put a'=0: $a' b+a+(123.456-a-b)(a'+1)+(-a'-1)(a+b)=0\to a+2b=123.456$ now we have a system of equation solving we get the pairs $(a,b,c)=(80.304,20.576,20.576),(0,61.728,61.728)$ .since the max,min of a is the same as for c, $max(c)+min(c)=80.304+0=\boxed{80.304}$ a visualization of the solutions:(the graph can show each of the solution is a min and max without having to take 2nd deravative)

Let $x=123.456 , y=3810.345984$ . Then we can find $a^2+b^2+c^2 + 2y=x^2$ $a^2 +b^2 +c^2 = x^2- 2y$ $a^2+b^2=x^2-2y-c^2 .... \boxed{1}$ $a+b=x-c .... \boxed{2}$

Then by using QM - AM inequalities : $\frac {\sqrt{a^2+b^2}}{\sqrt {2}} \ge \frac {a+b}{2}$ Substitute the first and the second equation then squaring both sides we get : $\frac {x^2-2y-c^2}{2} \ge \frac {x^2 +c^2 -2xc}{4}$ then solving it we get : $0 \ge 3c^2 -2xc +(4y-x^2)$ Two of these quadratic equation roots will be the minimum and the maximum value by solving the inequalities. So we just need to find the sum of the roots of this quadratic equation, by using vieta the sum of the root is : $\frac {2x}{3}= \frac {2 \times 123.456}{3} = 82.304$

[Here's the generalization. Don't ask why it's lengthy, as I'm not a good solution writer.]

Let $k>2$ be a positive real so that $a+b+c=k$ and $ab+bc+ca=(\frac { k }{ 2 } )^{ 2 }$ .

$=>a^2+b^2+c^2=\frac { 4{ k }^{ 2 }-2{ k }^{ 2 } }{ 4 } =\frac { { k }^{ 2 } }{ 2 } =2(ab+bc+ca)$ .

To find the minimum value of $c$ , assume that $a \ge b \ge c$ . It's obvious here that $b=c=0$ isn't possible.

Assume that $0 \ge b \ge c$ .

$=>b+c<k,bc \ge \frac { { k }^{ 2 } }{ 2 }$ (impossible)

Assume that $b \ge 0 \ge c$ .

$=>a+b=k-c,a^2+b^2=\frac { { k }^{ 2 } }{ 2 }-c^2$

$=>2(\frac { { k }^{ 2 } }{ 2 }-c^2) \ge (k-c)^2$

$=> c(2k-3c) \ge 0$ .

Observe that $k > 0, c \le 0 =>2k-3c>0=>c=0$ .

As $c < 0$ isn't possible, $min(c)=0$ .

To find the maximum value of $c$ , assume that $c \ge b \ge a > 0$ .

We have $a^2+b^2+c^2=2(ab+bc+ca)$

$=>(\sqrt { a } +\sqrt { b } +\sqrt { c } )(\sqrt { c } +\sqrt { b } -\sqrt { a } )(\sqrt { c } -\sqrt { b } +\sqrt { a } )(\sqrt { c } -\sqrt { b } -\sqrt { a } )=0$

It's obvious that $\sqrt { a } +\sqrt { b } +\sqrt { c }>0$ . Also, $\sqrt { c } +\sqrt { b } >\sqrt { a } +0$ and $\sqrt { a } +\sqrt { c } >\sqrt { b } +0$ .

$=>\sqrt { c } =\sqrt { b } +\sqrt { a }$

$=>c=a+b+2\sqrt{ab}$ .

Substitute this into the first equation, we have $a+b+\sqrt{ab}=\frac { k }{ 2 }$

$=>c=\sqrt{ab}+\frac { k }{ 2 }$ .

As we are looking for $max(c)$ , we need to find $max(\sqrt{ab})$ .

We have $b+a \ge 2\sqrt{ab}$

$=>4a+4b+4\sqrt{ab} \ge 3b+3c+6\sqrt{ab}$

$=>2k \ge 3b+3c+6\sqrt{ab} \ge 12\sqrt{ab}$

$=>\sqrt{ab} \le \frac { k }{ 6 }$

$=>c \le \frac { 2k }{ 3 }$ .

Minimum value of $c$ can be achieved when $b=a=\frac { k }{ 2 },c=0$ , while maximum value of c can be achieved when $c=\frac { 2k }{ 3 },a=b= \frac { k }{ 6 }$ .

Edit: $k>2$ may not be true. I am, still, not yet sure if I can write instead that $k>0$ .

If $a+b+c = 3k$ and $ab+ac+bc = 3d$ for some fixed $k, d$ , and $(a_0, b_0, c_0)$ is a solution, then $(2k-a_0, 2k-b_0, 2k-c_0)$ is also a solution (plug it in). Thus if the minimum value of $c$ is $c_m$ , the maximum value of $c$ must be $2k-c_m$ , and their sum is $2k$ . Plugging $3k = 123.456$ gives $2k = \boxed{82.304}$ .

Of course, this requires us to first show that there exists such solution, and the maximum and minimum of $c$ exist. But once you get both of these, the rest follows immediately.

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To show these two, note that $a,b,c$ are the roots of $x^3 - 3kx^2 + 3dx - e = 0$ for some real $e$ . Its derivative is $3x^2 - 6kx + 3d$ . If this is always positive, then it always has only one root for any $e$ , so there doesn't exist such solution. Thus first we must show that $3x^2 - 6kx + 3d$ can be non-positive; this can be proven by simply checking that the determinant $36(k^2-d)$ is non-negative. That is, if $d \le k^2$ , we will have a solution.

Now, it's easy to see that if $e$ is very big or very small, then there is only one root. So there's a bounded interval $I$ where $e$ must be in this interval in order for the equation to have three roots. And since $e$ is bounded, there is also a bounded interval $J$ in which if $x \not\in J$ then $x^3 - 3k^2 + 3dx \not\in I$ , which means no $e$ will give this root, which means $x$ is not part of a solution. So $a,b,c \in J$ , showing bounded-ness.

For the general case, we assume $a+b+c=p$ $ab+bc+ca=q$ Now, $b=p-c-a$ . So, $a(p-c-a)+(p-c-a)c+ca=q$ $\implies ap-ca-a^2+cp-c^2-ca+ca=q$ $\implies a^2+a(c-p)+(c^2-cp+q)=0$ If we want $a$ to have a real value, we need to impose the following condition. $(c-p)^2-4(c^2-cp+q) \geq 0$ We may want to use the completing square technique here. After that, we'd get $(3c-p)^2 \leq 4p^2-12q$ $\implies -\sqrt{4p^2-12q} \leq 3c-p \leq \sqrt{4p^2-12q}$ $\implies \frac{p}{3}-\frac{2}{3}\sqrt{p^2-3q} \leq c \leq \frac{p}{3}+\frac{2}{3}\sqrt{p^2-3q}$ Hence, the upper and the lower bounds represent the maximum and minimum values for $c$ . [These two values can then be used to get the corresponding real values of $a$ and $b$ . Also, for the upper and lower bounds of $c$ to be real, we need to impose the condition: $p^2 \geq 3q$ .]

Therefore, the sum of max and min values of $c$ is $\frac{2p}{3}$ For this particular problem, we have $p^2 > 3q$ . So, the answer is $\frac{2 \times 123.456}{3}=82.304$

The generalization can be found out by simply squaring the first equation, and then using Titu's lemma.....