Posting a Problem!!

We've $x+y+z=5…(1)$ $xy+yz+zx=3$ $\Rightarrow x^2+y^2+z^2=19$ $\Rightarrow x^2+y^2+(5-x-y)^2=19$ Solving further we get, $\Rightarrow 2y^2+y(2x-10)+(2x^2-10x+6)=10$ Since $y$ is real, determinant of this quadratic equation should be non-negative. $\Rightarrow 4x^2+100-40x-8(2x^2-10x+6) \geq 0$ $\Rightarrow 3x^2-10x-13 \leq 0$ $\Rightarrow (x+1)(x-\frac{13}{3}) \leq 0$ $\Rightarrow -1 \leq x \leq \frac{13}{3}$ Sum is $-1+\frac{13}{3}=\frac{10}{3}$ Hence answer is $\large 10+3=\boxed{13}$

$\textbf{NOTE:}$ You may have a shorter process. This is an unconventional way of solving problems, but, i like it. So, here goes:

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From the given relations, and by using Vieta's we construct a polynomial whose roots are $x,y,z$

$P(x)=t^{3}-5t^{2}+3t+p$ and $p$ is a constant.

Now we analyse it's derivative to find out the nature of the curve at different intervals. This is very important.

$P'(x)=3x^{2}-10x+p$

now, observing it's roots, $P'(x)≥0$ $\text{iff }x\in (-\infty,\frac{1}{3})\cup (3,+\infty)$ and is negative elsewhere.

$\textbf{Now}$ the positive derivative of $P(x)$ signifies the increasing nature of the curve of $P(x)$ .

So, $P(x)$ increases on the interval $(-\infty,\frac{1}{3})$ from $-\infty$ to $P(\frac{1}{3})=p+\frac{13}{27}$

and $P(x)$ decreases on the interval $(\frac{1}{3},3)$ from $P(\frac{1}{3})$ to $P(3)=p-9$

So, $p+\frac{13}{27}$ and $p-9$ signifies the $\textbf{local maximum and minimum}$ respectively.

For the roots to be real, (although it isn't said, but i assumed) ,

$p+\frac{13}{27}≥0$ and $p-9≤0$

$\implies\large{P(\frac{1}{3})≥0}$ and $P(3)≤0$

Now, an observation that solves the problem:

$\text{NOTE THAT:}$

$\large{P(-1)=p-9≤0} \text{ and } \large{P(\frac{13}{3})=p+\frac{13}{27}≥0}$

$\textbf{So, the polynomial} P(x) \textbf{changes it's signs in the intervals:}$

$\large{(-1,\frac{1}{3});(\frac{1}{3},3);(3,\frac{13}{3})}$

So, it is obvious that the roots must lie in the above intervals..

So, we have, for any root:

$\text{least possible value}=\large{-1}$

$\text{greatest possible value}=\large{\frac{13}{3}}$

Adding them, we get $\frac{10}{3}$

OUR ANSWER: $\LARGE{\boxed{13}}$

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A bonus of this solution is that, we can determine the intervals of any of the $3$ roots of the polynomial at once(though it isn't required). they go as follows:

$\large{-1≤x_{k_1}≤\frac{1}{3}≤x_{k_2}≤3≤x_{k_3}≤\frac{13}{3}}$

where $x_{k_i}$ are the roots

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yes, this was lengthy, but you cannot possible deny that this was a good solution. :D

@Parth Lohomi , what do you think?

Put y = z = t and then eliminate t getting 3x^2 - 10x - 13 = 0, which gives 10/3 as the sum of the two extreme real values of x, namely -1 and 13/3. The point to be noted here is that it is a convex function.

Let $z = 5 - x - y$ and substitute into the nonlinear equation:

$xy + (5-x-y)(x+y) = 3 \Rightarrow x^2 + (y-5)x + (y^2 -5y + 3) = 0 \Rightarrow x = \frac{(5-y) \pm \sqrt{(y-5)^2 - 4(1)(y^2-5y+3)}}{2} = \frac{(5-y) \pm \sqrt{-3(y-5/3)^2 + 192/9}}{2}$ .

The discriminant is a concave-down parabola with vertex $(\frac{5}{3},\frac{192}{9})$ , which means the largest and smallest values for $x$ compute to:

$x = \frac{(5-5/3) \pm \sqrt{192/9}}{2} = \frac{5}{3} \pm \frac{4\sqrt{3}}{3}$ ,

and whose sum is $(\frac{5}{3} + \frac{4\sqrt{3}}{3}) + (\frac{5}{3} - \frac{4\sqrt{3}}{3}) = \frac{10}{3} = \frac{m}{n} \Rightarrow m+n = 10+3=\boxed{13}.$

APPLY AM>GM for y and z. THEN in the inequality formed place the value of x in place of y and z as per the equations given. so. on L.H.S. it becomes 5-x and on R.H.S. , square root of x^2 - 5x +3 solve and get yur answer upvote the solution if lioke it.