Saudi Arabia booklet 2024

The given inequality can be written as $$2k\cdot \frac{(a^2+b^2+c^2)-(ab+bc+ca)}{(a+b+c{)}^2}\le \frac{\sqrt{a^2+b^2+c^2}-\sqrt{ab+bc+ca}}{\sqrt{a^2+b^2+c^2}}$$ $$2k\cdot \frac{(a^2+b^2+c^2)-(ab+bc+ca)}{(a+b+c{)}^2}\le \frac{(a^2+b^2+c^2)-(ab+bc+ca)}{\sqrt{a^2+b^2+c^2}((a^2+b^2+c^2)+(ab+bc+ca))}.$$ Note that $$(a^2+b^2+c^2)-(ab+bc+ca)=\frac{1}{2}\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]\ge 0$$ so $$2k\sqrt{a^2+b^2+c^2}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+bc+ca}\right)\le (a+b+c{)}^2.$$ Consider $a=b=1$ and $c\to 0^{+}$ (isosceles triangle with the base arbitrary small) then $\text{LHS}\to 2k\cdot \sqrt{2}(\sqrt{2}+1)$ and $\text{RHS}\to 4$. Thus $$k\le \frac{4}{2\sqrt{2}(\sqrt{2}+1)}=2-\sqrt{2}.$$ For $k=2-\sqrt{2}$ then we need to prove that $$(4-2\sqrt{2})\sqrt{a^2+b^2+c^2}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+bc+ca}\right)\le (a+b+c{)}^2.$$ Denote $x=\sqrt{a^2+b^2+c^2}$ and $y=\sqrt{ab+bc+ca}$ then the above can be written as $$(4-2\sqrt{2})x(x+y)\le x^2+2y^2$$ $$x^2+2(2+\sqrt{2})xy-(2+\sqrt{2}{)}^2{y}^2\le 0.$$ This equivalent to $x\le y\sqrt{2}$ or $a^2+b^2+c^2\le 2(ab+bc+ca)$. The last inequality is true for $a,b,c$ are sidelengths of triangle since it can be written as $$a(b+c-a)+b(c+a-b)+c(a+b-c)\ge 0.$$ Hence, $k_{\max }=2-\sqrt{2}$.