Find minimum of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$ Let $a, b,c$ be the lengths of the sides of a triangle such that $a+b+c=2$, find the minimum value of $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$$
I don't have many ideas for this problem, my attemps:
The minimum value is $2\sqrt{2}+\dfrac{4}{3}$, the equality occurs for $a=b=c=\dfrac{2}{3}$
By Cauchy–Schwarz inequality, we have: $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge \dfrac{1}{\sqrt{2}}(a+b+b+c+c+a)=2\sqrt{2}$$
Thus, it suffices to prove that $$ab+bc+ca\ge\dfrac{4}{3}$$
This cannot be true because $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}=\dfrac{4}{3}$
Does anyone have any ideas, please give me a hint

My second proof:
Denote the expression by $f$.
Clearly, we have $a, b, c < 1$. We have
$$\sqrt{a^2 + b^2} - \frac{a + b}{\sqrt 2}
= \frac{\frac{(a - b)^2}{2}}{\sqrt{a^2 + b^2} + \frac{a + b}{\sqrt 2}}
\ge \frac{\frac{(a - b)^2}{2}}{\sqrt{1^2 + 1^2} + \frac{1 + 1}{\sqrt 2}} \ge \frac{(a - b)^2}{6}$$
which results in
$$\sqrt{a^2 + b^2} \ge \frac{a + b}{\sqrt 2} + \frac{(a - b)^2}{6}.$$
Thus, we have
\begin{align*}
 f &\ge \frac{a + b + b + c + c + a}{\sqrt 2} + \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{6} + ab + bc + ca\\
 &= \sqrt 2 (a + b + c) + \frac13 (a + b + c)^2\\
 &= 2\sqrt 2 + \frac43.
\end{align*}
Also, when $a = b = c = 2/3$, we have $f = 2\sqrt{2} + 4/3$.
Thus, the minimum of $f$ is $2\sqrt 2 + \frac43$.
