Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ac$

Question

If $a, b, c$ are positive real numbers such that $a+b+c=3$, prove that $\sqrt{a} +\sqrt{b} +\sqrt{c} \ge ab+bc+ac$

My try: First I tried dividing both sides by $\sqrt{abc}$. Then I squared both sides. Afterwards...... no afterwards. I haven't managed to do much from here.

Answer 1

$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$

$\sum_{cyc}(a^2+2\sqrt{a})\ge 3\sum_{cyc}a\ge 9$