Almost Vieta's

Firstly, note that $a+b=10-c$ and $ab=25-c(a+b)=25-c(10-c)$ . Thus, we have

$\begin{aligned} 0 &\leq (a-b)^2\\ &=(a+b)^2-4ab\\ &=(10-c)^2-4(25-c(10-c))\\ &=c^2-20c+100-100+40c-4c^2\\ &=20c-3c^2\\ &=c(20-3c) \end{aligned}$

Thus, $0\leq c \leq \frac{20}{3}$ , so the sum of the minimum and maximum values of $c$ is $\frac{20}{3}$ . These values in $(a,b,c)$ occur at $(5,5,0)$ and $(\frac{5}{3}, \frac{5}{3}, \frac{20}{3})$ . Therefore, the answer is $20+3=\boxed{23}$ .

From the identity $(a+b+c)^2 = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ , and substituting the values given , we get $a^{2}+b^{2}+c^{2}=50 \\ \implies a^{2}+b^{2}=50-c^2 \\ Also, \hspace{0.25cm} a+b=10-c \\$ From Cauchy-Schwarz Inequality , $a^{2}+b^{2} \geq\frac{1}{2}(a+b)^{2}$ Putting the values of $a^{2}+b^{2}$ and $a+b$ , we get: $50-c^2\geq\frac{1}{2}(10-c)^2$ On simplifying we are left with $0\leq 3c^{2}-20c$ which gives $0\leq c\leq\frac{20}{3}$ yeilding maximum value as $\boxed{\frac{20}{3}}$

if you take the derivative of $x^3-10x^2+25x$ you get $3x^2-20x+25$ and the sum of the zeros is $\frac{20}{3}$ when added together gives you $\boxed{23}$