If $\lambda_1,\lambda_2$ , $\lambda_1<\lambda_2$ are roots of the equation $x^2-(3pq-p-q-2)x+(2p^2q^2-2p^2q-2pq^2+2pq-2p-2q+4)$ show that they satisfy
$\lambda_1<p(q-1)-1<pq-2<\lambda_2$ where $p,q$ are primes with $p<q$.
I tried:
One of $\lambda_1$ or $\lambda_2$ must be greater then $pq-2$. Otherwise if both $\lambda_1,\lambda_2\le pq-2$ then $2(pq-2)\ge\lambda_1+\lambda_2=3pq-p-q-2$
$\implies 3pq-p-q-2\le 2pq-4\implies pq-p-q+2\le 0\implies pq+2\le p+q$ which is false.
Thus one of $\lambda_1$ or $\lambda_2$ must be greater then $pq-2$.
Without loss of generality we assume that $\lambda_2>pq-2$-------------------(1).
Now since $p>1$ we have $p(q-1)-1<pq-2$----------(2).
From (1) and (2), $p(q-1)-1<pq-2<\lambda_2$-------------------(3)
We just need to show that $\lambda_1<p(q-1)-1$.
However I am stuck in showing that
$\lambda_1<p(q-1)-1$.
Can someone help me to show this inequality?
Let $f(x)=x^2-(3pq-p-q-2)x+2p^2q^2-2p^2q-2pq^2+2pq-2p-2q+4.$
Thus, should be $$f(p(q-1)-1)=(q-1)(-pq+3p+3)<0$$ and $$f(pq-2)=-p^2q-pq^2+6pq-4p-4q+4<0.$$ Can you end it now?
By the way, we need $p(q-3)>3,$ which gives that $q\geq5,$ otherwise the statement is wrong.