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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.

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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?

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  • $\begingroup$ This is the same way the Aczel's inequality is proved. $\endgroup$ – Darkrai Dec 23 '17 at 7:08
  • $\begingroup$ @Greninja Yes, I added a reference. $\endgroup$ – Robert Z Dec 23 '17 at 7:10
  • $\begingroup$ Thanks a lot :-) $\endgroup$ – Darkrai Dec 23 '17 at 7:12
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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.

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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).

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