Familiar, isn't it?

We can also use AM-GM inequality.

Since $x^2+y^2+xy \ge 3xy$ and similar for the other two expressions, we have:

$\begin{aligned} \sqrt{\color{#3D99F6}{x^2+y^2+xy}} + \sqrt{y^2+z^2+yz} + \sqrt{z^2+x^2+zx} & \ge \sqrt{\color{#3D99F6}{3xy}} + \sqrt{3yz} + \sqrt{3zx} \\ & \ge \sqrt{3}\left(3\sqrt [3]{xyz} \right) = 3\sqrt{3}\sqrt [3]{xyz} \end{aligned}$

Also,

$\begin{aligned} xy + yz + zx & \le 9xyz \\ \text{But} \quad xy + yz + zx & \ge 3 \left( \sqrt [3] {xyz} \right)^2 \\ \Rightarrow 9xyz & \ge 3 \left( \sqrt [3] {xyz} \right)^2 \\ \Rightarrow \sqrt [3] {xyz} & \ge \frac{1}{3} \end{aligned}$

Therefore,

$\begin{aligned} \sqrt{x^2+y^2+xy} + \sqrt{y^2+z^2+yz} + \sqrt{z^2+x^2+zx} & \ge 3\sqrt{3}\sqrt [3]{xyz} \\ & \ge 3 \sqrt{3} \times \frac{1}{3} = \boxed{\sqrt{3}} \end{aligned}$

From the condition, we get $9\geq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\frac{9}{x+y+z}\Rightarrow x+y+z\geq 1$ The problem can be written as $A= \sqrt{\bigg(x+\frac{y}{2}\bigg)^2+\frac{3y^2}{4}}+\sqrt{\bigg(y+\frac{z}{2}\bigg)^2+\frac{3z^2}{4}}+\sqrt{\bigg(z+\frac{x}{2}\bigg)^2+\frac{3x^2}{4}}$ So using Minkowski inequality we get $A\geq\sqrt{3}(x+y+z)\geq\sqrt{3}\approx 1.73$ The equality holds when $x=y=z=\frac{1}{3}$

Here is a solution using a combination of both AM-GM and Cauchy-Schwarz inequality.

Note that $9xyz \geq xy+yz+xz \geq 3(xyz)^{\frac{2}{3}}$ by AM-GM inequality.

Simplifying, we get

$(xyz)^{\frac{1}{3}} \geq \frac{1}{3}$

Note that

$\displaystyle \sum_{cyc} \sqrt{{x^{2}+y^{2}+xy}}$

$=\displaystyle \sum_{cyc} \sqrt{{(x^{2}+y^{2}+xy)(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})}}$

$\geq \displaystyle \sum_{cyc} (\frac{1}{\sqrt{3}}.x+\frac{1}{\sqrt{3}}.y+\frac{1}{\sqrt{3}}.\sqrt{xy})$ by Cauchy-Schwarz inequality

$=\frac{2}{\sqrt{3}}(x+y+z)+\frac{1}{\sqrt{3}}.(\sqrt{xy}+\sqrt{yz}+\sqrt{xz})$

$\geq \frac{2}{\sqrt{3}}.3.(xyz)^{\frac{1}{3}}+ \frac{1}{\sqrt{3}}.3.(xyz)^{\frac{1}{3}}$ by AM-GM inequality,

$\geq \frac{2}{\sqrt{3}}.3.\frac{1}{3}+ \frac{1}{\sqrt{3}}.3.\frac{1}{3}$ (from above)

$=\frac{3}{\sqrt{3}}$

$=\boxed{\sqrt{3}}$

Equality holds when $x^{2}=y^{2}=xy, y^{2}=z^{2}=yz, z^{2}=x^{2}=xz$ (condition for Cauchy-Schwarz inequality to hold, which simplifies to $x=y=z=\frac{1}{3}$ ). This happens to be the same condition for AM-GM inequality to hold, which we apply multiple times in our solution. We check also that $xy+yz+xz \leq 9xyz$ when $x=y=z=\frac{1}{3}$

We can apply vector into this problem.