It's not true for all $x$, take $x=\frac{1}{2}$ and $a=b=c=1$. I'm assuming you want to find the minimum $x$ then. I claim this is $x=1$, to do this note that $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$ so it suffices to show that $a^2+b^2+c^2\ge ab+bc+ac$. This is left as an exercise to the reader, it's easy with AM-GM.
Since equality occurs at $a=b=c$, we're done.
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