Aczel's inequality Can anyone here please explain how to notice Aczel's identity in a certain question. I mean in questions involving  inequalities, on understanding the inequality one gets a hint to apply a certain inequality e. g.  Cauchy Schwarz, Jensen's   inequality, Chebyshev's inequality, etc.  So how does a question involving Aczel's inequality be identified.   
Eg. In this question how does one get to know that this question uses Aczel's identity.
Suppose $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are real numbers such that $$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2.$$Prove that $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$
Note:  If someone wishes he/she can post the solution to this problem.
 A: The inequality can be proved by using a similar approach as for the Aczel's inequality. 
Consider the quadratic function
$$f(x)=\sum_{k=1}^n(a_kx-b_k)^2-(x-1)^2\\
=x^2(\sum_{k=1}^na_k^2-1)-2x(\sum_{k=1}^na_kb_k-1)+(\sum_{k=1}^nb_k^2-1).$$
The given condition, 
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2$$
implies that $f$ has no real roots. Since $f(1)\geq 0$ then $f(x)>0$ for all real $x$. 
Can you take it from here?
A: Let $1-\sum\limits_{i=1}^na_i^2>0$.
Hence, $1-\sum\limits_{i=1}^nb_i^2>0$ and we got a contradiction with the Aczel's inequality.
A: Suppose $$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2 \tag{A}$$
The quantity on the RHS is nonnegative so for signs to work out there are just two cases:
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1) > 0 \text{ and } (b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > 0\tag{1} $$ 
$$(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1) < 0 \text{ and } (b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) < 0\tag{2} $$ 
In case (1), we have immediately $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$.
In case (2) we have that $1^2 > a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2$ (and $1^2 > b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2$) so by Aczel's inequality 
$$(1-a_1 ^ 2 - a_2 ^ 2 - \cdots - a_n ^ 2)(1-b_1 ^ 2 - b_2 ^ 2 - \cdots - b_n ^ 2) \leq (1 - a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2$$ which contradicts (A).  
