$a,\ b,\ c,\ x,\ y\ are\ all\ real\ and \ge 0.$
$STEP\ I:$
$0 \le \sqrt{xy} \le \dfrac{x + y}{2} \implies 0 \le \dfrac{x + y}{2} - \sqrt{xy} \implies$
$0 \le \left(\dfrac{x + y}{2} - \sqrt{xy}\right)^2 = \left(\dfrac{x + y}{2}\right)^2 - 2\sqrt{xy}\left(\dfrac{x + y}{2}\right) + xy \implies$
$0 \le \left(\dfrac{x + y}{2}\right)^2 - (x + y)\sqrt{xy}+ xy \implies$
$(x + y)\sqrt{xy} \le \left(\dfrac{x + y}{2}\right)^2 + xy.$
$STEP\ II:$
$\dfrac{a^2 + b^2 + c^2 + 3(ab + bc + ac)}{2} = \dfrac{2a^2 + 2b^2 + 2c^2 + 6(ab + bc + ac)}{4} =$
$\dfrac{(a^2 + 2ab + b^2) + (a^2 + 2ac + c^2) + (b^2 + 2bc + c^2) + 4(ab + bc + ac)}{4} =$
$\left\{\left(\dfrac{a + b}{2}\right)^2 + ab\right\} + \left\{\left(\dfrac{a + c}{2}\right)^2 + ac\right\} +\left\{\left(\dfrac{b + c}{2}\right)^2 + bc\right\}.$
$STEP\ III:$
$By\ STEP\ I:$
$(a + b)\sqrt{ab} \le \left(\dfrac{a + b}{2}\right)^2 + ab,\ and$
$(a + c)\sqrt{ac} \le \left(\dfrac{a + c}{2}\right)^2 + ac,\ and$
$(b + c)\sqrt{bc} \le \left(\dfrac{b + c}{2}\right)^2 + bc.$
$STEP\ IV:$
$Combining\ STEPS\ II\ and\ III:$
$ (a + b)\sqrt{ab} + (a + c)\sqrt{ac} + (b + c)\sqrt{bc} \le \dfrac{a^2 + b^2 + c^2 + 3(ab + bc + ac)}{2}.$