China Girls' Mathematical Olympiad

Solution I

By the AM-GM inequality and the given condition, we have $$u\cdot \frac{v+w}{2}+v\cdot \frac{w+u}{2}+w\cdot \frac{u+v}{2}\ge u\sqrt{vw}+v\sqrt{wu}+w\sqrt{uv}\ge 1,$$ or $$uv+vw+wu\ge 1.$$ Since $$(u-v{)}^2+(v-w{)}^2+(w-u{)}^2\ge 0,u^2+v^2+w^2\ge uv+vw+wu\ge 1,$$ it follows that $$(u+v+w{)}^2=u^2+v^2+w^2+2(uv+vw+wu)\ge 3,$$ or $u+v+w\ge \sqrt{3}$. The equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$.

$$\begin{array}{rl}\frac{(u+v+w{)}^4}{9}& ={\left(\frac{u+v+w}{3}\right)}^3\cdot 3(u+v+w)\\ & \ge 3uvw(u+v+w)\\ & =(uvw+vwu+wuv)(u+v+w)\end{array}$$ $$\ge (u\sqrt{vw}+v\sqrt{wu}+w\sqrt{uv}{)}^2=1,$$ which implies that $u+v+w\ge \sqrt{3}$. The equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$.