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(D) By the Law of Cosines,$\cos P = \frac{q^2+r^2-p^2}{2qr} = \frac{q^2+r^2}{2qr} - \frac{p^2}{2qr}$. Since $q^2+r^2 \geq 2qr$ (by $AM \geq GM$),we ha

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$A, B$

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(D) By the Law of Cosines,$\cos P = \frac{q^2+r^2-p^2}{2qr} = \frac{q^2+r^2}{2qr} - \frac{p^2}{2qr}$. Since $q^2+r^2 \geq 2qr$ (by $AM \geq GM$),we have $\frac{q^2+r^2}{2qr} \geq 1$. Thus,$\cos P \geq 1 - \frac{p^2}{2qr}$. So,$(A)$ is correct.$(B)$ The inequality $(p+q) \cos R \geq (q-r) \cos P + (p-r) \cos Q$ can be rewritten as $(p \cos R + r \cos P) + (q \cos R + r \cos Q) \geq q \cos P + p \cos Q$. Using the projection formula,this simplifies to $q + p \geq r$,which is true by the triangle inequality. So,$(B)$ is correct.$(C)$ By the Law of Sines,$\frac{q+r}{p} = \frac{\sin Q + \sin R}{\sin P}$. Since $\sin Q + \sin R \geq 2 \sqrt{\sin Q \sin R}$,we have $\frac{q+r}{p} \geq 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$. Thus,$(C)$ is incorrect.$(D)$ The condition $\cos Q > \frac{p}{r}$ implies $\sin R \cos Q > \sin P$,which simplifies to $\sin P + \sin(R-Q) > 2 \sin P$,or $\sin(R-Q) > \sin P$. This is not necessarily true for all triangles where $p Hence,the correct statements are $(A)$ and $(B)$.

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