Inequality of Numerators

We see that the LHS simply equals $3(a+b+c)$ . Dividing both sides by $a+b+c$ , we have $3\le \dfrac{1}{k}$

Solving for $k$ : $k\le \dfrac{1}{3}$ so our answer is $1+3=\boxed{4}$ .

$\frac{a}{1}$ $+$ $\frac{b}{1}$ $+$ $\frac{c}{1}$ $+$ $\frac{a+b}{1}$ $+$ $\frac{b+c}{1}$ $+$ $\frac{c+a}{1}$ $≤$ $\frac{a+b+c}{k}$

$\frac{a+b+c+(a+b)+(b+c)+(c+a)}{1}$ $≤$ $\frac{a+b+c}{k}$

$\frac{3a+3b+3c}{1}$ $≤$ $\frac{a+b+c}{k}$

$3(a+b+c)$ $≤$ $\frac{a+b+c}{k}$

$3k$ $≤$ $\frac{a+b+c}{a+b+c}$

$3k$ $≤$ $1$

$k$ $≤$ $\frac{1}{3}$

so, $\frac{p}{q}$ = $\frac{1}{3}$

$p$ $+$ $q$ = $1$ $+$ $3$

$p$ $+$ $q$ = $4$

3(a+b+c)<=(a+b+c)/k Let x = a+b+c 3x = x/k k= 1/3 p+q = 4

(a+b+c)/k must be equal to 3(a+b+c). so the value of k should be '1/3'. Then 1+3=4

let us assume a=b=c=1;then evaluating the question we get 4