Symmetric Conditions But Not Symmetric Equality

We are trying to extremize the homogeneous function $Q \; = \; \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ subject to the homogenous constraint $7(a^2 + b^2 + c^2) \; = \; 11(ab + ac + bc)$ and so we can add the constraint $a + b + c \; = \;1$ without loss of generality. Thus we need to extremize $Q \; = \; \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$ subject to the conditions $a+b+c \; = \; 1 \qquad \qquad a^2 + b^2 + c^2 \; = \; \tfrac{11}{25} \;.$ Note that these constraints require $(a,b,c)$ to lie on a circle $\mathcal{C}$ in $\mathbb{R}^3$ which is perpendicular to the vector $(1\;1\;1)^T$ . This circle passes through the points $(\tfrac35,\tfrac15,\tfrac15)$ , $(\tfrac15,\tfrac35,\tfrac15)$ , $(\tfrac15,\tfrac15,\tfrac35)$ , and so has centre $(\tfrac13,\tfrac13,\tfrac13)$ . Since $\left(\begin{array}{c} \frac35 \\ \frac15 \\ \frac15 \end{array}\right) - \left(\begin{array}{c} \frac13 \\ \frac13 \\ \frac13 \end{array}\right) \; = \; \left(\begin{array}{c} \frac{4}{15} \\ -\frac{2}{15} \\ -\frac{2}{15}\end{array} \right) \qquad \qquad \left(\begin{array}{c} \frac{4}{15} \\ -\frac{2}{15} \\ -\frac{2}{15}\end{array} \right) \times \tfrac{1}{\sqrt{3}}\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right) \; = \; \left(\begin{array}{c} 0 \\ -\frac{2}{15}\sqrt{3} \\ \frac{2}{15}\sqrt{3}\end{array} \right)$ it follows that the coordinates of a general point on the circle $\mathcal{C}$ are $\begin{array}{rcl} (a,b,c) & = & \big(\tfrac13,\tfrac13,\tfrac13\big) + \tfrac{2}{15}\cos\theta(2,-1,-1) + \tfrac{2}{15}\sin\theta(0,-\sqrt{3},\sqrt{3}) \\ & = & \left(\tfrac13 + \tfrac{4}{15}\cos\theta,\tfrac13 + \tfrac{4}{15}\cos\big(\theta+\tfrac{2\pi}{3}\big),\tfrac13 + \tfrac{4}{15}\cos\big(\theta + \tfrac{4\pi}{3}\big)\right) \end{array}$ and the value of $Q$ at this general point on the circle $\mathcal{C}$ is (after a shed-load of algebra) $Q \; = \; \frac{270}{55 - \cos3\theta} - 3 \;.$ Thus it is clear that $\tfrac{51}{28} \; =\; \tfrac{270}{56} - 3 \; \le \; Q \; \le \; \tfrac{270}{54} - 3 \; = \; 2$ on the constraint circle $\mathcal{C}$ . Thus $\tfrac{51}{28}$ and $2$ are the minimum and maximum values of $Q$ subject to the given constraint, and since $\tfrac{51}{28} + 2 = \tfrac{107}{28}$ , we see that the answer is $\boxed{135}$ .

By choosing appropriate values of $\theta$ , we note that $Q$ is maximized when $a,b,c$ are equal to $\tfrac35,\tfrac15,\tfrac15$ in some order, and that $Q$ is minimized when $a,b,c$ are equal to $\tfrac{7}{15},\tfrac{7}{15},\tfrac{1}{15}$ in some order.

Relevant wiki: Vieta's Formula Problem Solving - Advanced

Since the numerator and denominator of each of the fractions for $S:= \dfrac a{b+c} + \dfrac b{a+c} + \dfrac c{a+b}$ has the same degree (of 1), we can normalize it by assuming that $a+b+c=1$ . Then $S= \dfrac a{1-a} + \dfrac b{1-b} + \dfrac c{1-c} = \dfrac{3abc + 2(ab+ac+bc) - (a+b+c)}{(abc) - (ab+ac+bc) + (a+b+c) - 1)} \qquad \qquad (1)$

We are given the constraint of $\begin{aligned} && a^2+b^2+c^2 = \dfrac{11}7 (ab+ac+bc) \\ && \Rightarrow (a+b+c)^2 - 2(ab+ac+bc) = \dfrac{11}7 (ab+ac+bc) \\ && \Rightarrow ab+ac+bc = \dfrac7{25} \; . \end{aligned}$

Let $f(x) = x^3 + Ax^2+Bx+C$ denote a cubic polynomial with roots $a,b$ and $c$ . Then by Vieta's formula , $A = -(a+b+c) = -1, B = ab+ac+bc = \dfrac7{25}$ and $C =-abc < 0$ because it is given that $a,b$ and $c$ are positive reals. Substituting these values gives $f(x) = x^3 -x^2 + \dfrac7{25} x + C$ .

Since all of its ( $f(x)$ ) roots are positive, then the cubic discriminant is non-negative:

$\Delta_3 = (-1)^2 \left( \dfrac7{25}\right)^2 - 4(1)\left( \dfrac7{25}\right)^3 - 27(1^2)(C^2) + 18(1)(-1)\left( \dfrac7{25}\right) (C) \geq 0$

Simplifying the inequality gives us $421875 C^2+16250 C+147 \leq 0$ , or equivalently $\left( C + \dfrac{49}{3375} \right) \left( C + \dfrac3{125} \right) \leq 0 \Leftrightarrow - \dfrac3{125} \leq C \leq -\dfrac{49}{3375} \; .$

Because $C = -abc$ , then $\dfrac{49}{3375} \leq abc \leq \dfrac3{125}$ .

Substitute all the relevant values into $(1)$ gives

$S = \dfrac{3abc + 2\left(7/25\right) - 1}{abc - 7/25 + 1 - 1} = -\dfrac{75abc - 11}{25abc - 7} = - \left( 3 + \dfrac{32}{25abc-7} \right) \; .$

Since the asymptote of $G(X) = - \left( 3 + \dfrac{32}{25X-7} \right)$ is at $X = \dfrac7{25}$ and we are only interested in the range of $\dfrac{49}{3375} \leq X \leq \dfrac3{125}$ , and $X = \dfrac7{25}$ falls outside of this domain, so $G(X)$ is defined in the domain $\left [ \dfrac{49}{3375} , \dfrac3{125} \right ]$ . Thus, there exists both a minimum value and a maximum value of $S$ .

We want to find the following values:

$\begin{aligned} \min S &=& \min \left[ - \left( 3 + \dfrac{32}{25abc-7} \right) \right ] = -\max \left(3 + \dfrac{32}{25abc-7} \right) = -3 -32 \max\left( \dfrac{1}{25abc-7} \right), \\ \max S &=& \max \left[ - \left( 3 + \dfrac{32}{25abc-7} \right) \right ] = -\min \left(3 + \dfrac{32}{25abc-7} \right) = -3 -32\min\left( \dfrac{1}{25abc-7} \right) \; . \end{aligned}$

Thus, $S$ is minimized when $abc$ is minimized; $S$ is maximized when $abc$ is maximized.

$\min S = -3 - 32 \left( \dfrac1{25 \left( \frac{49}{3375}\right) - 7} \right) = \dfrac{51}{28}, \qquad \max S = -3 - 32 \left( \dfrac1{25 \left( \frac{3}{125}\right) - 7} \right) = 2 \; .$

Now, we need to prove that there exist positive values of $a,b$ and $c$ such that the $\min S = \dfrac{51}{28}$ and $\max S = 2$ , separately.

When $\min S = \dfrac{51}{28}$ , $abc = \dfrac{49}{3375}$ , the values of $a,b$ and $c$ satisfy the polynomial $f(x) = x^3 - x^2 + \dfrac7{25} - \dfrac{49}{3375}$ , and its roots occur when $f(x) = 0$ :

$\begin{array} { r r l } & &x^3 - x^2 + \dfrac7{25} x - \dfrac{49}{3375} &= 0 \\ &\Rightarrow &3375x^3 -3375x^2 + 945x - 49 &= 0 \\ &\Rightarrow &(15x-7)^2(15x-1) &= 0 \\ &\Rightarrow &x &= \dfrac1{15}, \dfrac7{15}, \dfrac7{15} \; . \end{array}$

The last step above can be shown by applying rational root theorem .

Similarly, when $\max S = 2$ , $abc = \dfrac3{125}$ , the values of $a,b$ and $c$ satisfy the polynomial $f(x) = x^3 - x^2 + \dfrac7{25} - \dfrac{3}{125}$ , and its roots occur when $f(x) = 0$ :

$\begin{array} { r r l } & & x^3 - x^2 + \dfrac7{25} x - \dfrac{3}{125}& = 0 \\ &\Rightarrow & 125x^3 - 125x^2 + 35x-3 & = 0 \\ & \Rightarrow &(5x-3)(5x-1)^2 &= 0 \\ &\Rightarrow & x &= \dfrac15, \dfrac15, \dfrac35 \; . \end{array}$

Thus, the solutions to unordered triplets $(a,b,c)$ when $S$ is minimized/maximized are $\left( \dfrac15, \dfrac15, \dfrac35 \right)$ and $\left( \dfrac1{15}, \dfrac7{15}, \dfrac7{15} \right)$ .

But wait! Earlier, we have normalized the constraints by introducing the additional condition of $a+b+c= 1$ and we need to relax this constraint. So we need to revise our answer:

$\min S = \dfrac{51}{28}$ is achieved when the unordered triplets $(a,b,c)$ is $(k,k,3k)$ ,
$\max S = 2$ is achieved when the unordered triplets $(a,b,c)$ is $(k,7k,7k)$ ,
where $k$ is any positive constant.

Hence, $\min S + \max S = \dfrac{51}{28} + 2 = \dfrac{107}{28} \Rightarrow m+n = \boxed{135}$ .

Since the condition is homogenous, we can normalize it (by multiplying/dividing) each term by a constant $k$ such that $ab+bc+ca = 7$ . This gives us $a+b+c = 5$ .

Let's $c=\max\{a,b,c\}$ so we get $14-3(a+b) \geq 14 - 10 = 4 > 0$ . $\frac{3abc+(a+b+c)^{3}-2(a+b+c)(ab+bc+ca)}{(ab+bc+ca)(a+b+c)-abc}=\frac{3abc+55}{35-abc}$ $7=ab+bc+ca=c(a+b)+ab=(a+b)(5-a-b)+ab$ Let $a+b=s;ab=p$ . $abc=p(5-s)=[7-s(5-s)](5-s)$ We are going to prove $abc\geq \frac{49}{27}$ So that we can prove $abc\leq 3$ . And we can max and min,then the answer is 135