1. Use the cosine rule for Option A

In triangle $PQR$, $p^2=q^2+r^2-2qr\cos P.$ So, $\cos P=\frac{q^2+r^2-p^2}{2qr}.$ Now compare with $1-\frac{p^2}{2qr}=\frac{2qr-p^2}{2qr}.$ Thus, \cos P-\left(1-\frac{p^2}{2qr}\right)=\frac{q^2+r^2-p^2-(2qr-p^2)}{2qr}=rac{(q-r)^2}{2qr}\ge 0. Hence, $\cos P\ge 1-\frac{p^2}{2qr}.$ So A is true.

2. Check Option B

We use cosine rule: $\cos P=\frac{q^2+r^2-p^2}{2qr},\qquad \cos Q=\frac{p^2+r^2-q^2}{2pr},\qquad \cos R=\frac{p^2+q^2-r^2}{2pq}.$

We need to test: $\cos R\ge \left(\frac{q-r}{p+q}\right)\cos P+\left(\frac{p-r}{p+q}\right)\cos Q.$

Substitute the cosine-rule expressions. Let $S=\cos R-\left(\frac{q-r}{p+q}\right)\cos P-\left(\frac{p-r}{p+q}\right)\cos Q.$ After simplifying algebraically, this becomes $S=\frac{r(p+r)(q+r)-r^3-p^2q-pq^2}{2pqr}.$ A cleaner simplification is obtained by multiplying throughout by $2pq(p+q)$:

$2pq(p+q)S=(p+q)(p^2+q^2-r^2)-\frac{p(q-r)}{r}(q^2+r^2-p^2)-\frac{q(p-r)}{r}(p^2+r^2-q^2).$

On simplification, this reduces to $2pq(p+q)S=\frac{(p+q-r)\big(p^2+2pq+q^2+r(p+q)\big)}{r}.$ Since in a triangle, $p+q>r,$ and all sides are positive, the second factor is also positive. Therefore $S>0.$ Hence the required inequality holds: $\cos R\ge \left(\frac{q-r}{p+q}\right)\cos P+\left(\frac{p-r}{p+q}\right)\cos Q.$ So B is true.

3. Check Option C

The option states $\frac{q+r}{p}<2\frac{\sqrt{\sin q\sin R}}{\sin P}.$ This is not even a meaningful standard triangle expression because $q$ is a side length, while $R$ and $P$ are angles. The term $\sin q$ is dimensionally inconsistent in the usual triangle notation, where $q$ denotes a side, not an angle.

So as written, C is false / invalid.

4. Check Option D

Given $p<q$ and $p<r$, we test whether $\cos Q>\frac{p}{r},\qquad \cos R>\frac{p}{q}.$

Take a counterexample: let $p=3,\quad q=4,\quad r=5.$ This is a valid triangle and indeed $p<q$, $p<r$.

Now, $\cos Q=\frac{p^2+r^2-q^2}{2pr}=\frac{9+25-16}{2\cdot 3\cdot 5}=\frac{18}{30}=0.6,$ while $\frac{p}{r}=\frac{3}{5}=0.6.$ So $\cos Q>\frac{p}{r}$ is false.

Also, $\cos R=\frac{p^2+q^2-r^2}{2pq}=\frac{9+16-25}{2\cdot 3\cdot 4}=0,$ while $\frac{p}{q}=\frac{3}{4}=0.75.$ Thus $\cos R>\frac{p}{q}$ is also false.

Hence D is false.

5. Final conclusion

The true statements are: $\boxed{A,\ B}$

This matches the stored correct answer.