3. Determine the maximum possible value of the expression

$$
27 a b c+a \sqrt{a^{2}+2 b c}+b \sqrt{b^{2}+2 c a}+c \sqrt{c^{2}+2 a b}
$$

where $a, b, c$ are positive real numbers such that $a+b+c=\frac{1}{\sqrt{3}}$.
Solution. First, it is observed that when $a=b=c=\frac{1}{3 \sqrt{3}}$, the value of the expression is $\frac{2}{3 \sqrt{3}}$, which suggests conjecturing that

$$
27 a b c+a \sqrt{a^{2}+2 b c}+b \sqrt{b^{2}+2 c a}+c \sqrt{c^{2}+2 a b} \leq \frac{2}{3 \sqrt{3}}
$$

To prove the conjecture, one can apply the Cauchy-Schwarz inequality, $(\vec{u} \cdot \vec{v})^{2} \leq$ $\|\vec{u}\|^{2}\|\vec{v}\|^{2}$, to the vectors $\vec{u}=(\sqrt{a}, \sqrt{b}, \sqrt{c})$ and $\vec{v}=(\sqrt{b c}, \sqrt{c a}, \sqrt{a b})$, yielding

$$
9 a b c=(\sqrt{a b c}+\sqrt{a b c}+\sqrt{a b c})^{2} \leq(a+b+c)(b c+c a+a b)
$$

Multiplying both sides of the above inequality by 3 and considering the constraint, we get

$$
27 a b c \leq 3(a+b+c)(b c+c a+a b)=\sqrt{3}(b c+c a+a b)
$$

On the other hand, since $a \sqrt{a^{2}+2 b c}=\frac{1}{\sqrt{3}} \sqrt{3 a^{2}\left(a^{2}+2 b c\right)}$, applying the arithmetic mean-geometric mean inequality to the numbers $3 a^{2}$ and $a^{2}+2 b c$ yields

$$
a \sqrt{a^{2}+2 b c}=\frac{1}{\sqrt{3}} \sqrt{3 a^{2}\left(a^{2}+2 b c\right)} \leq \frac{1}{\sqrt{3}} \frac{3 a^{2}+\left(a^{2}+2 b c\right)}{2}=\frac{1}{\sqrt{3}}\left(2 a^{2}+b c\right)
$$

Similarly, we have

$$
b \sqrt{b^{2}+2 c a} \leq \frac{1}{\sqrt{3}}\left(2 b^{2}+c a\right)
$$

and

$$
c \sqrt{c^{2}+2 a b} \leq \frac{1}{\sqrt{3}}\left(2 c^{2}+a b\right)
$$

Combining the above inequalities and considering the constraint again, we obtain

$$
\begin{aligned}
& 27 a b c+a \sqrt{a^{2}+2 b c}+b \sqrt{b^{2}+2 c a}+c \sqrt{c^{2}+2 a b} \\
\leq & \sqrt{3}(b c+c a+a b)+\frac{1}{\sqrt{3}}\left(2 a^{2}+b c+2 b^{2}+c a+2 c^{2}+a b\right) \\
= & \frac{2}{\sqrt{3}}\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)=\frac{2}{\sqrt{3}}(a+b+c)^{2}=\frac{2}{3 \sqrt{3}}
\end{aligned}
$$

This proves the conjecture and the maximum value of the expression is $\frac{2}{3 \sqrt{3}}$.