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How to prove $(a+b+c)^2 > 3(bc+ca+ab)$ when a, b, c are positive, unequal and of course, real.

The only thing I have been able to do is:

$a^2+b^2+c^2 > 0$

So $(a+b+c)^2 - 2 (ab+bc+ac) >0$

$(a+b+c)^2 > 2(ab+bc+ac) $

I need one more $(ab+bc+ac)$ on the right-hand side, but I don't see any way to do it. I tried using A.M. - G.M. inequality, but even after trying many terms, I failed t obtain it.

I would prefer an answer without Cauchy-Schwarz inequality, using only A.M. - G.M. inequality if it is required.

Swapnil Rustagi
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4 Answers4

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Hint: add $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$ and $c^2+a^2\ge 2ca$ together.

Eclipse Sun
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  • Reason I found it to be best answer: I only had to correct the first statement of MY ATTEMPT, so I had to begin with $2 (a^2 + b^2 + c^2) > 2 (ab + bc + ca)$, and the rest follows. Also, this is a more direct method, and since it uses A.M-G.M. inequality I like this better. – Swapnil Rustagi Nov 30 '14 at 06:41
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$$ (a+b+c)^2 - 3(ab + bc + ac) = a^2+b^2+c^2-ab-bc-ac=\\ =\frac{1}{2}(a^2 -2ab+b^2) + \frac{1}{2}(b^2 -2bc+c^2) + \frac{1}{2}(a^2 -2ac+c^2)=\\ \frac{1}{2}\left((a-b)^2+(b-c)^2+(a-c)^2\right) \geq 0. $$

Jihad
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Notice $2 ( a^2 + b^2 + b^2 - ab -bc - ac ) = (a-b)^2 + (a-c)^2 + (b-c)^2 \geq 0 $

Hence,

$$ a^2 + b^2 + c^2 \geq ab + bc + ac $$

Now, use this and the fact that

$$ (a+b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc+ac) $$

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For $$W= {\rm AB}+ {\rm C}$$ $$W= -\,{\rm AD}+ {\rm E}$$ $${\rm B},\,{\rm C},\,{\rm D},\,{\rm E}\geqq 0\,\therefore\,W= \frac{{\rm BE}+ {\rm CD}}{{\rm B}+ {\rm D}}\geqq 0 \tag{29}$$ Such as $$W= \sum\,a^{\,2}- \sum\,ab= \left ( a- c \right )^{\,2}+ \left ( a- b \right )\left ( c- b \right )\geqq 0$$ $$W= \sum\,a^{\,2}- \sum\,ab= \left ( a+ c- 2\,b \right )^{\,2}+ 3\left ( a- b \right )\left ( b- c \right )\geqq 0$$ $$\therefore\,W= \frac{3(\,a- c\,)^{\,2}+ (\,a+ c- 2\,b\,)^{\,2}}{4}\geqq 0$$ I call it $\lceil$ DRIVE!S.O.S $\rfloor$