<s>[INST] 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. 
 [/INST] \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.
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