Prove the inequality $a^2 + b^2 +c^2 \ge ab +bc +ac$

Question

How do I prove the inequality \begin{equation*} a^2 + b^2 +c^2 \ge ab +bc +ac \end{equation*} where $ a,b,c\in\mathbb{R} $ and $a,b,c>0$?

I obtained only $(a+b+c)^2\ge 3(ab+bc+ac)$ and some other variations

Answer 1

Note that $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right).$$

Answer 2

multiply by $2$ ,then you have : $$2a^2+2b^2+2c^2 ≥ 2ab+2bc+2ac\\ 2a^2+2b^2+2c^2 -( 2ab+2bc+2ac) ≥0\\(a^2+b^2-2ab)+(b^2+c^2-2bc)+(a^2+c^2-2ac)≥ 0\\$$

Answer 3

Hint We have $x^2+y^2 \geq 2 x y .$

Write down the three inequialities for $a, b$ and $a,c$ and $b,c$ and then add them.

Answer 4

Another proof is: Consider $f(a) = a^2 -(b+c)a + b^2+c^2 -bc$ as a function of variable $a$. Observe that this is a quadratic polynomial with leading coefficient $1 > 0$. So all that you need to show is the discriminant $\triangle < 0$. Indeed, $\triangle = B^2-4AC = (b+c)^2 - 4(b^2+c^2-bc) = b^2+2bc+c^2 - 4b^2-4c^2+4bc = -3b^2-3c^2+6bc = -3(b-c)^2 \leq 0$. Thus $f(a) \geq0, \forall a \in \mathbb{R}$, and this implies the result...