Consider a triangle PQR having sides of lengths p, q and r opposite...

(A) Applying cosine rule

$cosP = \dfrac{{q}^{2} + {r}^{2} - {p}^{2}}{2qr}$

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

We know, using $A . M . \ge G . M .$

\left(\dfrac{q}{r} + \dfrac{r}{q}\right) \ge 2$</p><p>$⇒ \dfrac{1}{2}\left(\dfrac{q}{r} + \dfrac{r}{q}\right) \ge 1$</p><p>$⇒ \dfrac{1}{2}\left(\dfrac{q}{r} + \dfrac{r}{q}\right)\dfrac{- {p}^{2}}{2qr} \ge 1 - \dfrac{{p}^{2}}{2qr}$</p><p>$⇒ cosP \ge 1 - \dfrac{{p}^{2}}{2qr}$</p><p>(B) \left(p + q\right)cosR \ge \left(q - r\right)cosP + \left(p - r\right)cosQ$

\left(pcosR + rcosP\right) + \left(qcosR + rcosQ\right) \ge \left(qcosP + pcosQ\right)$</p><p>using projection formula</p><p>$q + p \ge r$</p><p>Which is true as sum of two sides of triangles is greater than third side.</p><p>(C)$\dfrac{q + r}{p} < \dfrac{2\sqrt{sinQsinR}}{sinP}$</p><p>$\dfrac{q + r}{p} < 2\sqrt{\dfrac{sinQsinR}{\sin^{2}P}}$</p><p>Applying sine rule</p><p>$⇒ \dfrac{q + r}{p} < 2\sqrt{\dfrac{qr}{{p}^{2}}}$</p><p>$⇒ \dfrac{q + r}{2p} < \sqrt{\dfrac{qr}{{p}^{2}}}$</p><p>Applying$A . M . \ge G . M .$</p><p>$\dfrac{\dfrac{q}{p} + \dfrac{r}{p}}{2} \ge \sqrt{\dfrac{qr}{{p}^{2}}}$</p><p>So, (C) is not true</p><p>$⇒ \dfrac{q + r}{2p} \ge \sqrt{\dfrac{qr}{{p}^{2}}}$</p><p>(D)$p < q&p < r$</p><p>$⇒ p$is smallest side</p><p>Let$cosQ > \dfrac{p}{r}$and$cosR > \dfrac{p}{q}$is true</p><p>$⇒ p < rcosQ$</p><p>$p < qcosR$</p><p>adding both inequalities</p><p>$⇒ 2p < rcosQ + qcosR$</p><p>using projection formula</p><p>$⇒ 2p < p$</p><p>$2 < 1

Which is not correct.