# 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.

• @ILOVEMATH Got any tips? – Samuli Lehtonen Sep 12 '13 at 17:36
• – Martin Sleziak Jan 30 '15 at 8:10

\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}

• 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.

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

\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.

• Hello, welcome to Math.SE. Your answer is rather hard to read. For some basic information about writing maths at this site see e.g. here, here, here and here. Your answer also looks like it was copy-pasted. Please be sure to give attribution where it's due. – Lord_Farin Oct 23 '13 at 12:54

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