Simple AM-GM inequality

Question

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?

Answer 1

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

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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.

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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?