# Simple AM-GM inequality

Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that $$a+b+c \geq ab+bc+ca$$ I was able to prove that \begin{align} a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\ &\geq \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2}= \\ &= ab+bc+ca \end{align}

but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?

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Update: Upps, that is the same as that of @Vincent, sorry didn't see that first. However, it's a bit more explicte

A nicer way to use your results and proceed frm there is the following. You've got that $$a^2+b^2+c^2 \ge ab+bc+ca$$ Now use the fact that $(a+b+c)=S=3$ and multiply each side of your original inequality by S this is $$(a+b+c)S \ge S(ab+bc+ca)$$ or, (where we use now that $S=3$) $$(a+b+c)(a+b+c) \ge 3(ab+bc+ca)$$ Then $$a^2+b^2+c^2 + 2(ab+bc+ca) \ge 3(ab+bc+ca)$$ and $$a^2+b^2+c^2 \ge ab+bc+ca$$ which is exactly that, what you've already proven on your own.

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It was that easy! I didn't see it, thank you. – Emmet Feb 27 '13 at 10:30
Ok, but now that I think about it... How to proceed from that? Any idea? – Emmet Feb 27 '13 at 10:39
hmm, after that you've done the work, and you can proceed to the cafeteria... ;-) – Gottfried Helms Feb 27 '13 at 15:22

$$9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca)$$

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Thank you very much. – Emmet Feb 27 '13 at 10:30

Hint: Multiply the original inequality by $a+b+c$ on the LHS and $3$ on the RHS, expand and eliminate common terms and you will arrive at something you have proved.

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Thank you very much. – Emmet Feb 27 '13 at 10:30

$9 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \geq 3(ab+bc + ac)$

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Thank you very much. – Emmet Feb 27 '13 at 10:32

Reformulate first $a+b+c=S$ and $a=M+d \qquad b=M-d$ then

$$S \ge M^2-d^2 + (S-2M)2M$$ reorganize $$3M^2-2SM+S +d^2 \ge 0$$

Now make use of the given definition that $S=3$. We get $$3(M^2-2M+1) +d^2 \ge 0$$ $$3(M-1)^2 +d^2 \ge 0$$ which is always true.

Well, this focuses "when and how" it makes sense to introduce the condition that $S=3$. Unfortunately the step with the AM-GM-inequality is lost. But maybe you can combine your steps with this derivations?

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