Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$? Can I prove it like this: Let's say that $a=b=c$ so we get "If $a \geq 0$ then $3a^2 ≥ 3a^2$" Now I take the negation of that statement and get "If $a \geq 0$ then $3a^2 < 3a^2$" The anti-thesis is obviously wrong which makes the original thesis right? Is this a correct way to do this, if not can you give some tips for what to do. I am not looking for complete solution as these are my homework and I really need to practice. 
 A: Hint: We have equality when $a=b=c$. So perhaps express the difference in terms of $a-b$, $b-c$, and $c-a$.
Remark: One cannot expect to be able to prove that something holds for all $a,b,c$ by looking only at a special case. And one cannot expect to prove a mathematical result by logical manipulation: the informal mathematical idea comes first, with logic playing a supporting role. 
A: Hint: Notice that $(a-b)^2 \geq 0.$ What happens when you sum three different such  inequalities: $(a-b)^2 \geq 0$, $(b-c)^2 \geq 0$, $(a-c)^2 \geq 0$?
A: \begin{align}
& a^2+b^2+c^2-ab-bc-ca\geq0 \\
&\iff \frac12(2a^2+2b^2+2c^2-2ab-2bc-2ca)\geq0 \\
&\iff \frac12((a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2))\geq0\\
&\iff \frac12((a-b)^2+(b-c)^2+(c-a)^2)\geq0\\
&\iff (a-b)^2+(b-c)^2+(c-a)^2\geq0\\
\end{align}
A: \begin{align}
a^2 + b^2 + c^2 - ab - bc - ca
&= (1/2)(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)\\
&= (1/2)(a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2)\\
&= (1/2)[(a-b)^2 + (b-c)^2 + (c-a)^2]
\end{align}
Now we know that, the square of a number is always greater than or equal to zero.
Hence, $(1/2)[(a-b)^2 + (b-c)^2 + (c-a)^2] ≥ 0 \implies$ it is always non-negative.
