## Task A-2.4.

Determine the largest real constant $\lambda$ such that for all positive real numbers $u$, $v, w$ for which $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geq 1$, the inequality $u+v+w \geq \lambda$ holds.
## Solution.

We claim that the maximum possible value of the constant $\lambda$ is $\sqrt{3}$.

Indeed, it is easy to see that for $u=v=w=\frac{\sqrt{3}}{3}$, we have $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$ and $u+v+w=\sqrt{3}$. Therefore, the maximum value of the constant $\lambda$ is less than or equal to $\sqrt{3}$.

We will show that for $u, v, w > 0$ such that $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geq 1$, it holds that $u+v+w \geq \sqrt{3}$.

First method. From the given conditions and the inequality between the arithmetic and geometric means, we have

$$
u \cdot \frac{v+w}{2}+v \cdot \frac{w+u}{2}+w \cdot \frac{u+v}{2} \geq u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geq 1
$$

which implies

$$
u v+v w+w u \geq 1
$$

Since

$$
\begin{aligned}
& (u-v)^{2}+(v-w)^{2}+(w-u)^{2} \geq 0 \\
& u^{2}+v^{2}+w^{2} \geq u v+v w+w u \geq 1
\end{aligned}
$$

it follows that

$$
(u+v+w)^{2}=u^{2}+v^{2}+w^{2}+2(u v+v w+w u) \geq 3
$$

or $u+v+w \geq \sqrt{3}$. Equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$.

Second method. By applying the arithmetic-geometric mean inequality and then the Cauchy-Schwarz inequality, we have

$$
\begin{aligned}
& \frac{(u+v+w)^{4}}{9}=\left(\frac{u+v+w}{3}\right)^{3} \cdot 3(u+v+w) \geq 3 u v w(u+v+w)= \\
& =(u v w+v w u+w u v)(u+v+w) \geq(u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v})^{2} \geq 1
\end{aligned}
$$

Thus, $u+v+w \geq \sqrt{3}$. Equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$.

Third method. Let $x=\sqrt{u v}, y=\sqrt{v w}, z=\sqrt{w u}$. Then, due to the given condition, we have

$$
x y+y z+z x=v \sqrt{w u}+w \sqrt{u v}+u \sqrt{v w} \geq 1
$$

Note that $u=\frac{z x}{y}, v=\frac{x y}{z}, w=\frac{y z}{x}$. By applying the inequality between the arithmetic and geometric means, we have

$$
2(u+v+w)=\left(\frac{z x}{y}+\frac{x y}{z}\right)+\left(\frac{x y}{z}+\frac{y z}{x}\right)+\left(\frac{y z}{x}+\frac{z x}{y}\right) \geq 2 x+2 y+2 z
$$

or $u+v+w \geq x+y+z$. As in the first solution, we get

$$
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \geq 3(x y+y z+z x) \geq 3
$$

and finally

$$
(u+v+w)^{2} \geq(x+y+z)^{2} \geq 3
$$

or $u+v+w \geq \sqrt{3}$. Equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$.