Find minimum of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$

Question

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

Answer 1

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 $2\sqrt 2 + \frac43$.

Answer 2

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 $2\sqrt 2 + \frac43$.