# Double Inequality. $ab+bc+ca \leq \frac{1}{8}\sum{\sqrt{(1-ab)(1-bc)}} \leq a^2+b^2+c^2$

Let $a,b,c$ be non-negative numbers such that $a+b+c=1.$ Prove that :

$$ab+bc+ca \leq \frac{1}{8}\sum_{cyc}{\sqrt{(1-ab)(1-bc)}} \leq a^2+b^2+c^2.$$

Thanks:)

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Are we summing over cyclic permutations of $(a,b,c)$? – Alex Becker Sep 14 '12 at 17:17

To start, let's note that $$1 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \tag 1$$

### First Inequality

By means of Cauchy-Schwarz inequality, we can write $$8(ab + bc + ca) = \sum_{cyc} \Bigg(\frac a {\sqrt 2} \cdot \frac b {\sqrt 2} + \frac b {\sqrt 2} \cdot \frac c {\sqrt 2} + \frac c {\sqrt 2} \cdot \frac a {\sqrt 2} + \\ \sqrt {2ab} \cdot \sqrt {2ab} + \sqrt {2bc} \cdot \sqrt {2bc} + \sqrt {2ca} \cdot \sqrt {2ca} + \frac c {\sqrt 2} \cdot \frac a {\sqrt 2} \Bigg) \leq \\ \sum_{cyc} \sqrt {\frac {a^2 + b^2 + c^2} 2 + 2ab + 2bc + 2ca + \frac {c^2} 2} \cdot \sqrt {\frac {a^2 + b^2 + c^2} 2 + 2ab + 2bc + 2ca + \frac {a^2} 2} =: I$$ With identity $(1)$ holding and being $ab \leq (a^2 + b^2)/2$ (and cyclics), we get $$I \leq \sum_{cyc} \sqrt {a^2 + b^2 + c^2 + ab + 2bc + 2ca} \cdot \sqrt {a^2 + b^2 + c^2 + 2ab + bc + 2ca} = \\ \sum_{cyc} \sqrt{(1 - ab)(1 - bc)}$$

### Second Inequality

Using Cauchy-Schwarz inequality we get $$\sum_{cyc} \sqrt{(1 - ab)} \cdot \sqrt{(1 - bc)} \leq 3 - (ab + bc + ca) =: I$$ With another application of CS inequality and of the identity $(1)$, we arrive to $$I = 3(a^2 + b^2 +c^2) + 5(a\cdot b + b\cdot c + c\cdot a) \leq\\ 3(a^2 + b^2 + c^2) + 5(a^2 + b^2 + c^2) =\\ 8(a^2 + b^2 + c^2)$$

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You are amazing. I like a lot how you solve the inequality, but I have a question :) How did you think to decompose that sum? When I saw it , immediately I said it is wrong but I stayed with my friend and took the proof step by step and we couldn't believe one like this, you were right:) Nice, very nice solution – Iuli Sep 17 '12 at 8:05
@Iuli Thanks a lot! It was funny reading your comment. – AlbertH Sep 17 '12 at 13:45
Please don't forget me. How did you think to write $\displaystyle 8(ab+bc+ca)$ like this: $$\sum_{cyc} \Bigg(\frac a {\sqrt 2} \cdot \frac b {\sqrt 2} + \frac b {\sqrt 2} \cdot \frac c {\sqrt 2} + \frac c {\sqrt 2} \cdot \frac a {\sqrt 2} + \\ \sqrt {2ab} \cdot \sqrt {2ab} + \sqrt {2bc} \cdot \sqrt {2bc} + \sqrt {2ca} \cdot \sqrt {2ca} + \frac c {\sqrt 2} \cdot \frac a {\sqrt 2}\Bigg)$$ Thanks a lot:) – Iuli Sep 17 '12 at 15:37