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 : 

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

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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 
$$
I = 3(a^2 + b^2 +c^2) + 5(a\cdot b + b\cdot c + c\cdot a) \leq\\ 3(a^2 + b^2 + c^2) + 5(a^2 + b^2 + c^2) =\\ 8(a^2 + b^2 + c^2)
$$
