I claim that $ (u + v + w)^2 \ge 3(u\sqrt {vw} + v\sqrt {uw} + w\sqrt {uv})$ Proof: $ (u + v + w)^2 = u^2 + v^2 + w^2 + 2uv + 2uw + 2vw \ge 3uv + 3uw + 3vw$ $ = 3(u\frac {v + w}{2} + v\frac {u + w}{2} + w\frac {u + v}{2})\ge 3(u\sqrt {vw} + v\sqrt {uw} + w\sqrt {uv})$ Hence, $ u + v + w \ge \sqrt {3(u\sqrt {vw} + v\sqrt {uw} + w\sqrt {uv})}\ge \sqrt {3}$. We can attain equality when $ u = w = v = \frac {\sqrt{3}}{3}$

By Muirhead, $\frac{1}{2}[2,0,0]\succ \frac{1}{2}\left[1,\frac{1}{2},\frac{1}{2}\right]$ and $[2,2,0]\succ \left[1,\frac{1}{2},\frac{1}{2}\right]$. Adding these two inequalities yields $u^2+v^2+w^2+2uv+2vw+2wu\ge 3u\sqrt{vw}+3v\sqrt{wu}+3w\sqrt{uv}$. This implies $(u+v+w)^2\ge 3(u\sqrt{vw}+v\sqrt{wu}+w\sqrt{uv})\ge 3\implies u+v+w\ge \sqrt{3}$, as Rofler noted.

April wrote: Let $ u, v, w$ be positive real numbers such that $ u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv} \geq 1$. Find the smallest value of $ u + v + w$. The stronger one is Let $u,v,w\in\mathbb R_+$ satisfy $u\sqrt{uv}+v\sqrt{vw}+w\sqrt{wu}\geq 1,$ then we have $u+v+w\geq \sqrt 3.$ It follows from Vasc's inequality.

By rearrangement, \[(u+v+w)^2 = u^2 + v^2 + w^2 + 2(uv + uw + vw) \ge 3(u\sqrt{vw}+v\sqrt{wu}+w\sqrt{uv}) \ge 3\] Hence the minimum value is $\sqrt{3}$.

Replacement:\[u\sqrt{vw}=a;v\sqrt{wu}=b;w\sqrt{uv}=c\] Then:\[u=\frac{a^2}{\sqrt{abc}};v=\frac{b^2}{\sqrt{abc}};w=\frac{c^2}{\sqrt{abc}};a+b+c\geq{1}\] \[\frac{a^2+b^2+c^2}{\sqrt{abc}}\geq\frac{\frac{1}{3}(a+b+c)^2}{\sqrt{abc}}\geq\frac{\frac{1}{3}(a+b+c)^2}{\sqrt{\frac{(a+b+c)^3}{27}}}=\sqrt{3(a+b+c)}\geq{\sqrt{3}}\] \[min=\sqrt{3}\] \[ u=v=w=\frac{\sqrt{3}}{3}\]

2016 San Diego Math Olympiad

$gt<=>\sqrt{uvw}(\sqrt{u}+\sqrt{v}+\sqrt{w})\geq 1<=>1\leq (\sqrt{u}+\sqrt{v}+\sqrt{w})^2uvw\leq 3(u+v+w)(u+v+w)^3\frac{1}{27}=(\sum u)^4\frac{1}{9}=>\sum u\geq \sqrt{\sqrt{9}}=\sqrt{3}$ We can attain equality when: $ u = w = v = \frac {\sqrt{3}}{3}$

Cleaner solution: Let $u=a^2$, $v=b^2$ and $w=c^2$. Then the given inequality becomes $abc(a+b+c)\geq 1$ Using AM-GM gives $\bigg(\frac{a+b+c}{3}\bigg)^4\geq\frac{1}{3}$ Using another power mean law, this time between AM and the RMS. $\bigg(\frac{a^2+b^2+c^2}{3}\bigg)^2\geq\frac{1}{3}$ Thus becomes $a^2+b^2+c^2\geq\sqrt{3}$. Substitute back to get $u+v+w\geq\sqrt{3}$. Equality is indeed achieved at $u=v=w$ as can easily be checked.

Let $x=\sqrt{uv}, y=\sqrt{vw}, z=\sqrt{wu}$. Now, $u\sqrt{vw}+v\sqrt{wu}+w\sqrt{vu}=xy+yz+zx \geq 1$. Also, we can see that $u=\frac{zx}{y}, v=\frac{xy}{z}, w=\frac{yz}{x}$. From $AM-GM$ inequality we see that $2(u+v+w)=(\frac{zx}{y}+\frac{xy}{z})+(\frac{xy}{z}+\frac{yz}{x})+(\frac{yz}{x}+\frac{zx}{y}) \geq 2(x+y+z)$. So, $u+v+w \geq x+y+z$ Since $uv+vw+wu \geq 1$ and $u^2+v^2+w^2 \geq uv+vw+wu \geq 1$ we have $(u+v+w)^2=u^2+v^2+w^2+2(uv+vw+wu) \geq 3$ or $u+v+w \geq \sqrt{3}$. Equality is when $u=v=w=\frac{\sqrt{3}}{3}$.

Since we know that $(u+v+w)^2 \ge 3(uw+vu+wu)$, as well as $ uv + vw + wu \ge u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv}$, we can combine these two to get our minimum value as root 3.

Let $ u, v, w$ be positive real numbers such that $ u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv} \geq 1.$ Prove that$$ u + v + w+\sqrt[3]{uvw}\geq \frac{4}{\sqrt 3}$$

mynameisbob12 wrote: Since we know that $(u+v+w)^2 \ge 3(uw+vu+wu)$, as well as $ uv + vw + wu \ge u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv}$, we can combine these two to get our minimum value as root 3. Of course

sqing wrote: Let $ u, v, w$ be positive real numbers such that $ u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv} \geq 1.$ Prove that$$ u + v + w+\sqrt[3]{uvw}\geq \frac{4}{\sqrt 3}$$