Consider a triangle P Q R having sides of lengths p , q and r opposite to the angles P , Q and R ,…

(A) Applying cosine rule

$\cos P=\frac{q^2+r^2-p^2}{2qr}$

$=\frac{1}{2}\left(\frac{q}{r}+\frac{r}{q}\right)-\frac{p^2}{2qr}$

We know, using $A.M.\ge G.M.$

$\left(\frac{q}{r}+\frac{r}{q}\right)\ge 2$

$\Rightarrow \frac{1}{2}\left(\frac{q}{r}+\frac{r}{q}\right)\ge 1$

$\Rightarrow \frac{1}{2}\left(\frac{q}{r}+\frac{r}{q}\right)\frac{-p^2}{2qr}\ge 1-\frac{p^2}{2qr}$

$\Rightarrow \cos P\ge 1-\frac{p^2}{2qr}$

(B) $\left(p+q\right)\cos R\ge \left(q-r\right)\cos P+\left(p-r\right)\cos Q$

$\left(p\cos R+r\cos P\right)+\left(q\cos R+r\cos Q\right)\ge \left(q\cos P+p\cos Q\right)$

using projection formula

$q+p\ge r$

Which is true as sum of two sides of triangles is greater than third side.

(C) $\frac{q+r}{p}<\frac{2\sqrt{\sin Q\sin R}}{\sin P}$

$\frac{q+r}{p}<2\sqrt{\frac{\sin Q\sin R}{{\sin }^{2}P}}$

Applying sine rule

$\Rightarrow \frac{q+r}{p}<2\sqrt{\frac{qr}{p^2}}$

$\Rightarrow \frac{q+r}{2p}<\sqrt{\frac{qr}{p^2}}$

Applying $A.M.\ge G.M.$

$\frac{\frac{q}{p}+\frac{r}{p}}{2}\ge \sqrt{\frac{qr}{p^2}}$

So, (C) is not true

$\Rightarrow \frac{q+r}{2p}\ge \sqrt{\frac{qr}{p^2}}$

(D) $p<q$ & $p<r$

$\Rightarrow p$ is smallest side

Let $\cos Q>\frac{p}{r}$ and $\cos R>\frac{p}{q}$ is true

$\Rightarrow p<r\cos Q$

$p<q\cos R$

adding both inequalities

$\Rightarrow 2p<r\cos Q+q\cos R$

using projection formula

$\Rightarrow 2p<p$

$2<1$

Which is not correct.