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
 A: 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 $2\sqrt 2 + \frac43$.
A: Denote the expression by $f$.
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
 &(\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2})^2\\
 =\,& 2a^2 + 2b^2 + 2c^2
 + 2\sqrt{(a^2 + b^2)(c^2 + b^2)}
 + 2\sqrt{(c^2 + b^2)(c^2 + a^2)}\\
 &\quad
 + 2\sqrt{(c^2 + a^2)(b^2 + a^2)}\\
 \ge\,& 2a^2 + 2b^2 + 2c^2 + 2(ac + b^2) + 2(c^2 + ab) + 2(bc + a^2)\\
 =\,& 4a^2 + 4b^2 + 4c^2 + 2ab + 2bc + 2ca.
\end{align*}
Thus, we have
$$f \ge \sqrt{4a^2 + 4b^2 + 4c^2 + 2ab + 2bc + 2ca} + ab + bc + ca. \tag{1}$$
Let $a = x + y, b = y + z, c = z + x$
for $x, y, z > 0$ (the so-called Ravi's substitution).
Then $x + y + z = 1$.
From (1), we have
\begin{align*}
 f &\ge \sqrt{10x^2 + 10y^2 + 10z^2 + 14xy + 14yz + 14zx}\\
&\quad + x^2 + y^2 + z^2 + 3xy + 3yz + 3zx\\
 &= \sqrt{10(x + y + z)^2 - 6(xy + yz + zx)} + (x + y + z)^2 + xy + yz + zx\\
 &= \sqrt{10 - 6q} + 1 + q \tag{2}
\end{align*}
where $q = xy + yz + zx$.
Since $x + y + z = 1$ and $x, y, z> 0$, we have $0 < xy + yz + zx \le 1/3$. Thus, $0 < q \le 1/3$.
It is easy to prove that $\sqrt{10 - 6q} + 1 + q \ge \sqrt{8} + \frac43$.
From (2), we have $f \ge  \sqrt{8} + \frac43$.
Also, when $a = b = c = 2/3$, we have $f = \sqrt{8} + 4/3$.
Thus, the minimum of $f$ is $2\sqrt 2 + \frac43$.
