# 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.

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@ILOVEMATH Got any tips? –  Samuli Lehtonen Sep 12 '13 at 17:36

\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\\ \end{align}

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You missed the last nice step, $(a-b)^2+(b-c)^2+(c-a)^2$! One should observe the OPs expression is symmetric in $a,b,c$ and attains equality when $a=b=c$ so this motivates your solution strongly. –  Pedro Tamaroff Sep 12 '13 at 17:45
@PeterTamaroff Yes, that was on purpose. But I'll add that. –  Alraxite Sep 12 '13 at 17:48
It was somehow extremely hard for me to recognize that you can put the equation into that form but I get it now. I guess I just need to practice more so I start seeing those things. –  Samuli Lehtonen Sep 12 '13 at 17:55

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.

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Hint: Notice that $(a-b)^2 \geq 0.$ What happens when you sum three different such inequalities: $(a-b)^2 \geq 0$, $(b-c)^2 \geq 0$, $(a-c)^2 \geq 0$?