Inequalities Involving Trigonometric Ratios and Sides of Triangle

To determine which of the assertions are TRUE, let us analyze each option individually:

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Option (A):

By the Cosine Rule for $\cos P$: $\cos P = \frac{q^2 + r^2 - p^2}{2qr} = \frac{q^2 + r^2}{2qr} - \frac{p^2}{2qr}$

Since $(q - r)^2 \ge 0$ for any real numbers $q$ and $r$: $q^2 + r^2 \ge 2qr \implies \frac{q^2 + r^2}{2qr} \ge 1$

Substituting this inequality back into the expression for $\cos P$: $\cos P \ge 1 - \frac{p^2}{2qr}$

Thus, Option (A) is TRUE.

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Option (B):

Consider the given inequality: $\cos R \ge \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$

Since $p+q > 0$, multiply both sides by $(p+q)$: $(p+q)\cos R \ge (q-r)\cos P + (p-r)\cos Q$

Expanding and rearranging the terms: $p\cos R + q\cos R \ge q\cos P - r\cos P + p\cos Q - r\cos Q$ $(p\cos R + r\cos P) + (q\cos R + r\cos Q) \ge p\cos Q + q\cos P$

By the Projection Formulas for any triangle $PQR$: $p\cos R + r\cos P = q$ $q\cos R + r\cos Q = p$ $p\cos Q + q\cos P = r$

Substituting these formulas into the rearranged inequality: $q + p \ge r \implies p + q \ge r$

By the triangle inequality, the sum of any two side lengths is strictly greater than the third side ($p + q > r$), so $p + q \ge r$ is always true.

Thus, Option (B) is TRUE.

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Option (C):

By the Sine Rule, $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_c$, where $R_c$ is the circumradius: $\frac{q+r}{p} = \frac{\sin Q + \sin R}{\sin P}$

For any triangle, $\sin Q > 0$ and $\sin R > 0$. Applying the AM-GM inequality to $\sin Q$ and $\sin R$: $\frac{\sin Q + \sin R}{2} \ge \sqrt{\sin Q \sin R} \implies \sin Q + \sin R \ge 2\sqrt{\sin Q \sin R}$

Dividing both sides by $\sin P > 0$: $\frac{q+r}{p} = \frac{\sin Q + \sin R}{\sin P} \ge 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$

Therefore, the assertion $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$ is incorrect.

Thus, Option (C) is FALSE.

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Option (D):

Consider an isosceles triangle with side lengths $p = 1$, $q = 3$, and $r = 3$. Here, $p < q$ ($1 < 3$) and $p < r$ ($1 < 3$) are satisfied.

Using the Cosine Rule to find $\cos Q$: $\cos Q = \frac{p^2 + r^2 - q^2}{2pr} = \frac{1^2 + 3^2 - 3^2}{2(1)(3)} = \frac{1}{6}$

Now evaluating $\frac{p}{r}$: $\frac{p}{r} = \frac{1}{3}$

Comparing the two values: $\cos Q = \frac{1}{6} < \frac{1}{3} = \frac{p}{r}$

This contradicts the assertion $\cos Q > \frac{p}{r}$.

Thus, Option (D) is FALSE.

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Conclusion:

The correct choices are A and B.