23rd Korean Mathematical Olympiad Final Round

Put $BC=a$, $CA=b$, $AB=c$, $QR=p$, $RP=q$, $PQ=r$ and let $AP=x$, $BQ=y$, $CR=z$. For short, Since $x+y=c$, $y+z=a$, $z+x=b$, we have $$x=s-a,y=s-b,z=s-c\left(s=\frac{a+b+c}{2}\right).$$ The Cosine Law applied to the triangles $ABC$ and $ARQ$ thus yields $$a^2=b^2+c^2-2bc\cos A=(b-c{)}^2+2bc(1-\cos A)$$ and $$p^2=2x^2(1-\cos A)=2(s-a{)}^2(1-\cos A).$$ Now, canceling out $1-\cos A$ from the above two identities, we may represent $p^2$ in terms of $a,b,c$ as $$\begin{array}{rl}{p}^2& =(s-a{)}^2\cdot \frac{a^2-(b-c{)}^2}{bc}\\ & =\frac{4(s-a)(s-b)(s-c)}{abc}\cdot a(s-a).\end{array}(1)$$ Meanwhile, note that $$4(s-a)(s-b)=(b-c-a)(a-b-c)=c^2-(b-a{)}^2\le c^2$$ and similarly that $$4(s-b)(s-c)\le a^2\text{and}4(s-c)(s-a)\le b^2.$$ This yields $8(s-a)(s-b)(s-c)\le abc$, which in turn, together with (1), yields $p^2\le \frac{a(s-a)}{2}$, or equivalently, $${\left(\frac{a}{p}\right)}^3\ge 2\sqrt{2}{\left(\frac{a}{s-a}\right)}^{3/2}.$$ The parallel arguments yield $${\left(\frac{b}{q}\right)}^3\ge 2\sqrt{2}{\left(\frac{b}{s-b}\right)}^{3/2}\text{and}{\left(\frac{c}{r}\right)}^3\ge 2\sqrt{2}{\left(\frac{c}{s-c}\right)}^{3/2}.$$ Thus, denoting by $M$ the left hand side of the given inequality, we obtain $$\begin{array}{rl}M& \ge 2\sqrt{2}\left\{{\left(\frac{a}{s-a}\right)}^{3/2}+{\left(\frac{b}{s-b}\right)}^{3/2}+{\left(\frac{c}{s-c}\right)}^{3/2}\right\}\\ & \ge \frac{2\sqrt{2}}{\sqrt{3}}{\left(\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}\right)}^{3/2}\end{array}(2)$$ where the second inequality holds by Jensen's inequality. Moreover, since $$a\ge b\ge c⟺\frac{1}{s-a}\ge \frac{1}{s-b}\ge \frac{1}{s-c}.$$ we have by the Chevyshev inequality and the AM-GM inequality $$\begin{array}{rl}\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}& \ge \frac{1}{3}(a+b+c)\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\\ & \ge \frac{a+b+c}{\{(s-a)(s-b)(s-c){\}}^{1/3}}\\ & =\frac{(a+b+c)s^{1/3}}{\{s(s-a)(s-b)(s-c){\}}^{1/3}}\\ & =\frac{1}{2^{1/3}}{\left(\frac{L^2}{T}\right)}^{2/3}\end{array}(3)$$ Now, we conclude from (2) and (3) $$M\ge \frac{2\sqrt{2}}{\sqrt{3}}\cdot \frac{1}{\sqrt{2}}\frac{L^2}{T}=\frac{2}{\sqrt{3}}\cdot \frac{L^2}{T}.$$ as required.

□