# $\sum a+b+c \sum \frac{ab+bc+ca}{a+b+c} \sum \frac{abc}{ab+bc+ca} \leq \sum a \sum b \sum c$

Let $a_i,b_i,c_i$ be $>0$ ($1 \leq i \leq n$). Then we have $$\sum (a_i+b_i+c_i) \sum \frac{a_i b_i + b_i c_i + c_i a_i}{a_i+b_i+c_i} \sum \frac{a_i b_i c_i}{a_i b_i+b_i c_i + c_i a_i} \leq \sum a_i \sum b_i \sum c_i$$

This is problem #68 in Hardy, Polya and Littlewood's Inequality. It can be proved by using convexity (see e.g. http://www.win.tue.nl/~gwoegi/papers/cauchy.pdf) but I would like to see an elementary proof of this inequality. The problem is found in the chapter which deals with elementary means and associated inequality (such as arithmetic-geometric mean, Hölder and Minkowski's inequalities, ...). So there should be an elementary derivation of this inequality from some other well-known result. Of course we can rewrite the proof which relies on convexity but this is cheating. For example it is easy to prove by induction that $$\sum \frac{a_i b_i + b_i c_i + c_i a_i}{a_i+b_i+c_i} \leq \frac{\sum a_i \sum b_i+ \sum b_i \sum c_i + \sum c_i \sum a_i}{\sum a_i+b_i+c_i}$$ and $$\sum \frac{a_i b_i c_i}{a_i b_i+b_i c_i+c_i a_i} \leq \frac{\sum a_i \sum b_i \sum c_i}{\sum a_i \sum b_i+ \sum b_i \sum c_i + \sum c_i \sum a_i}$$ so we get the result, but we knew these inequalities were true in the first place thanks to convexity.

Lemma 1: $$\sum_{i=1}^{n}(a_{i}+b_{i})\sum_{i=1}^{n}\dfrac{a_{i}b_{i}}{a_{i}+b_{i}}\le\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i} \tag{1}$$ proof: $$\Longleftrightarrow \sum_{i=1}^{n}\left(a_{i}+b_{i}-\dfrac{4a_{i}b_{i}}{a_{i}+b_{i}}\right)\ge\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}-\dfrac{4\displaystyle\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}}{\displaystyle\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}}$$ $$\Longleftrightarrow \sum_{i=1}^{n}\dfrac{(a_{i}-b_{i})^2}{a_{i}+b_{i}}\ge\dfrac{(\displaystyle\sum_{i=1}^{n}(a_{i}-b_{i}))^2}{\displaystyle\sum_{i=1}^{n}(a_{i}+b_{i})}$$ It's easy use of $Cauchy-Schwarz$ inequality!

Lemma 2: $$\sum_{i=1}^{n}(a_{i}+b_{i}+c_{i})\sum_{i=1}^{n}\dfrac{a_{i}b_{i}+a_{i}c_{i}+b_{i}c_{i}}{a_{i}+b_{i}+c_{i}}\le\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}+\sum_{i=1}^{n}b_{i}\sum_{i=1}^{n}c_{i}+\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}c_{i} \tag{2}$$

proof:
use Lemma 1, we have $$\sum_{i=1}^{n}((a_{i}+b_{i})+c_{i})\sum_{i=1}^{n}\dfrac{(a_{i}+b_{i})c_{i}}{a_{i}+b_{i}+c_{i}}\le(\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i})\sum_{i=1}^{n}c_{i}$$

and same as other two inequality, and add this three inequality by done!

Lemma 3: $$\left(\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}+\sum_{i=1}^{n}c_{i}\sum_{i=1}^{n}b_{i}+\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}c_{i}\right)\sum_{i=1}^{n}\dfrac{a_{i}b_{i}c_{i}}{a_{i}b_{i}+a_{i}c_{i}+b_{i}c_{i}}\le \sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\sum_{i=1}^{n}c_{i} \tag{3}$$

proof:
let $t_{i}=\dfrac{c_{i}b_{i}}{b_{i}+c_{i}}$, and use Lemma 1, we have $$\sum_{i=1}^{n}(a_{i}+t_{i})\sum_{i=1}^{n}\dfrac{a_{i}t_{i}}{a_{i}+t_{i}}\le\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}t_{i}$$ and $$\sum_{i=1}^{n}(c_{i}+b_{i})\sum_{i=1}^{n}\dfrac{c_{i}b_{i}}{c_{i}+b_{i}}\le\sum_{i=1}^{n}c_{i}\sum_{i=1}^{n}b_{i}$$ Now It's easy prove it!

and you can $(1)\times (2)\times (3)$, that equality prove it by done!

• That's exactly what I was looking for! – timofei Jun 4 '13 at 20:30