fuzzylogic wrote:

eh, if conclusion is assumed false, then either the first or the second of what you said is true, not necessarily both.

al.M.V. wrote: fuzzylogic wrote:

eh, if conclusion is assumed false, then either the first or the second of what you said is true, not necessarily both. If only one is true, then the L.H.S would be negative, whereas the R.H.S is positive, hence either neither or both have to be true.

Another solution: Suppose instead that $ a_{1}^2 + ... + a_{n}^2 \leq 1$ and $ b_{1}^2 + ... + b_{n}^2 \leq 1$. We shall prove that $ (a_{1}^2 + ... + a_{n}^2- 1)(b_{1}^2 + ... + b_{n}^2 - 1) \leq (a_{1}b_{1} + ... + a_{n}b_{n} - 1)^2$. Noting that Cauchy-Schwarz is valid for any, (not necessarily positive) reals, we have: $ (a_{1}b_{1} + ... + a_{n}b_{n})^2 \leq (a_{1}^2 + ... + a_{n}^2)(b_{1}^2 + ... + b_{n}^2) \leq 1$, so $ a_{1}b_{1} + ... + a_{n}b_{n} \leq |a_{1}b_{1} + ... + a_{n}b_{n}| \leq 1$. Let $ X = a_{1}^2 + ... + a_{n}^2, Y = b_{1}^2 + ... + b_{n}^2$, so $ 0 \leq X,Y \leq 1$. Since $ 0 \leq X+Y - 2\sqrt {XY}$, we have $ XY - (X+Y) \leq XY - 2 \sqrt {XY}$, so $ (X-1)(Y-1) \leq (\sqrt{XY} - 1)^2$, (1) But Cauchy-Schwarz gives us $ a_{1}b_{1} + ... + a_{n}b_{n} \leq \sqrt {XY}$, so since $ f(x) = (x-1)^2$ is decreasing for $ x \leq 1$, $ ( \sqrt {XY} - 1)^2 \leq (a_{1}b_{1} + ... + a_{n}b_{n} - 1)^2$ (2). Combining (1) and (2) gives us our desired inequality.

Let $A = a_1^2 + a_2^2 + \cdots + a_n^2$, $B = b_1^2 + b_2^2 + \cdots + b_n^2$, and $C = a_1b_1 + a_2 b_2 + \cdots + a_n b_n$. We are given that $(A-1)(B-1) > (C-1)^2$, which can be rearranged to $AB - C^2 > A+B - 2C$. Rewrite this inequality as $(\sqrt{AB} - C)(\sqrt{AB}+C) > (\sqrt{A} - \sqrt{B})^2 + 2(\sqrt{AB} - C)$, so $(\sqrt{AB}-C)(\sqrt{AB}+C-2) > (\sqrt{A} - \sqrt{B})^2 \geq 0$. By the Cauchy-Schwarz inequality, $\sqrt{AB} \geq C$ so we need $\sqrt{AB}+C-2 \geq 0$. But $C \leq \sqrt{AB}$, so $2 \sqrt{AB} \geq 2 \implies AB \geq 1$. Since $(A-1)(B-1) > 0$, either both $A<1,B<1$ or both $A>1,B>1$. In the former case, we have $AB < 1$, a contradiction, so we must have $A,B>1$, as desired.

I'm sorry, but what is Aczel's Inequality?

Aczel's Inequality states that if $A^2>a_1^2+a_2^2+\cdots+a_n^2$ and $B_2>b_1^2+b_2^2+\cdots+b_n^2$, then $\left(A^2-a_1^2-a_2^2-\cdots-a_n^2\right)\left(B^2-b_1^2-b_2^2-\cdots-b_n^2\right)\le$ $\left(AB-a_1b_1-a_2b_2-\cdots-a_nb_n\right)$. See the page about it on AoPSWiki.

Let $\mathbf{a}=\begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}$. Then \begin{align*} ||\mathbf{a}-\mathbf{b}||^2||\mathbf{a}||^2&=||\mathbf{a}||^2||\mathbf{b}||^2+||\mathbf{a}-\mathbf{b}||^2||\mathbf{a}||^2-||\mathbf{a}||^2||\mathbf{b}||^2\\ &=||\mathbf{a}||^2||\mathbf{b}||^2+\left(||\mathbf{a}-\mathbf{b}||^2-||\mathbf{b}||^2\right)||\mathbf{a}||^2\\ &=||\mathbf{a}||^2||\mathbf{b}||^2+\left\langle\mathbf{a}-2\mathbf{b},\mathbf{a}\right\rangle\cdot\left\langle\mathbf{a},\mathbf{a}\right\rangle\\ &=||\mathbf{a}||^2||\mathbf{b}||^2+\left\langle\mathbf{a}-\mathbf{b},\mathbf{a}\right\rangle^2-\left\langle\mathbf{b},\mathbf{a}\right\rangle^2\\ &\geq||\mathbf{a}||^2||\mathbf{b}||^2-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2\\ &=\left(||\mathbf{a}||^2-1\right)\left(||\mathbf{b}||^2-1\right)-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2+||\mathbf{a}||^2+||\mathbf{b}||^2-1\\ &>\left(\left\langle\mathbf{a},\mathbf{b}\right\rangle-1\right)^2-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2+||\mathbf{a}||^2+||\mathbf{b}||^2-1\\ &=||\mathbf{a}||^2-2\left\langle\mathbf{a},\mathbf{b}\right\rangle+||\mathbf{b}||^2\\ &=||\mathbf{a}-\mathbf{b}||^2 \end{align*}so $||\mathbf{a}||>1$. Similarly $||\mathbf{b}||>1$. Thus $a_1^2+\ldots+a_n^2>1$ and $b_1^2+\ldots+b_n^2>1$.

http://www.artofproblemsolving.com/Wiki/index.php/Aczel's_Inequality. kill by Aczel inequality

Let $a=\sum_{i=1}^{n} a_i^2 , b=\sum_{i=1}^{n} b_i^2 , c=\sum_{i=1}^{n} a_ib_i$. Assume for contradiction that $a<1$ and $b<1$. Given inequality can be written as $c^2 - 2c + (a + b - ab) < 0$. The function $f(x)=x^2 - 2x + (a + b - ab)$ has $2$ positive roots one of which is less than $1$ and that root is $1-\sqrt{(a-1)(b-1)}$. So $c > 1 - \sqrt{(a-1)(b-1)}$. By Cauchy we know that $ab\geq c^2 > (a-1)(b-1) + 1 - 2\sqrt{(a-1)(b-1)} \implies a + b + 2\sqrt{(a-1(b-1)} > 2$. But it can easily be proven by computing derivatives that when $0<a<1$ and $0<b<1$, $a + b + 2\sqrt{(a-1)(b-1)}$ takes its maximum value $2$ when $a=b$. Hence contradiction.

Apparently, essentially the same as an early problem from 2000 Mongolian Olympiad, Teacher-secondary level. See here.

there's a nice proof of aczel's inequality by constructing yet another quadratic function

Aczel's Inequality with $A=B=1$.