USA IMO

If exactly one of $a_1^2 + a_2^2 + \dots + a_n^2$ and $b_1^2 + b_2^2 + \dots + b_n^2$ is no greater than $1$, then the left-hand side of the inequality is no greater than $0$, and inequality $(*)$ is trivial. Now we assume that both are no greater than $1$. Set $a = 1 - a_1^2 - a_2^2 - \dots - a_n^2$ and $b = 1 - b_1^2 - b_2^2 - \dots - b_n^2$. Then both $a$ and $b$ are nonnegative real numbers. Multiplying both sides of the inequality $(*)$ by $4$ gives
$$(2 - 2a_1b_1 - 2a_2b_2 - \dots - 2a_n b_n)^2 \geq 4ab$$
Note that
$$\begin{aligned} & 2 - 2a_1b_1 - 2a_2b_2 - \cdots - 2a_n b_n \\ &= (a_1 - b_1)^2 + (a_2 - b_2)^2 + \cdots + (a_n - b_n)^2 + a + b \\ &\geq a + b \geq 0. \end{aligned}$$
It follows that
$$(2 - 2a_1b_1 - 2a_2b_2 - \cdots - 2a_n b_n)^2 \geq (a+b)^2 \geq 4ab,$$
as desired.

We use the following inequality of Aczel: If $x_1, \dots, x_m, y_1, \dots, y_m \in \mathbb{R}$ such that $x_1^2 > x_2^2 + \dots + x_m^2$, then
$$(x_1y_1 - x_2y_2 - \dots - x_m y_m)^2 \geq (x_1^2 - x_2^2 - \dots - x_m^2)(y_1^2 - y_2^2 - \dots - y_m^2).$$
To prove Aczel's Inequality, we consider the quadratic function
$$f(t) = (x_1 t + y_1)^2 - \sum_{i=2}^{m} (x_i t + y_i)^2$$
and note that $f(-y_1/x_1) \leq 0$. It follows that the discriminant is nonnegative, hence proving the desired inequality. Now we proceed to solve the problem at hand. It is clear that $a_1^2 + \dots + a_n^2 - 1$ and $b_1^2 + \dots + b_n^2 - 1$ have the same sign. If both are negative, then Aczel's Inequality yields
$$(1 - a_1 b_1 - \dots - a_n b_n)^2 \geq (1 - a_1^2 - \dots - a_n^2)(1 - b_1^2 - \dots - b_n^2),$$
contradicting the given inequality.

Define the vectors $\mathbf{u} = [a_1, a_2, \dots, a_n]$ and $\mathbf{v} = [b_1, b_2, \dots, b_n]$, and place $\mathbf{u}$ and $\mathbf{v}$ tail by tail at the origin $O$ to form two sides of a triangle in $\mathbb{R}^n$ by letting $A$ and $B$ be the points with the coordinates of $\mathbf{u}$ and $\mathbf{v}$. The given inequality is equivalent to $(\mathbf{u} \cdot \mathbf{u} - 1)(\mathbf{v} \cdot \mathbf{v} - 1) > (\mathbf{u} \cdot \mathbf{v} - 1)^2$, where “$\cdot$” denotes the dot product of two vectors. Expanding both sides and rearranging terms yields
$$(\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) - \mathbf{u} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v} > (\mathbf{u} \cdot \mathbf{v})^2 - 2(\mathbf{u} \cdot \mathbf{v}),$$
or
$$(\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) - (\mathbf{u} \cdot \mathbf{v})^2 > \mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} - 2(\mathbf{u} \cdot \mathbf{v}).$$
Factoring the right-hand side of the above inequality gives
$$(\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) - (\mathbf{u} \cdot \mathbf{v})^2 > (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}).$$
By the vector form of the Law of Cosines, we have $(\mathbf{u} \cdot \mathbf{v})^2 = (\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) \cos^2 \angle AOB$. Hence
$$(\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v})(1 - \cos^2 \angle AOB) > (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}),$$
or
$$(\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) \sin^2 \angle AOB > (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v});$$
that is, $OA^2 \cdot OB^2 \sin^2 \angle AOB > AB^2$. Hence
$$[AOB] = \frac{1}{2} OA \cdot OB^2 \sin \angle AOB > \frac{1}{2} \cdot AB.$$
Therefore, the length of the altitude from $O$ to $AB$ must be greater than $1$. Yet the altitude is the shortest segment connecting $O$ to $AB$, and so $1 < OA^2 = \mathbf{u} \cdot \mathbf{u} = \sum_{i=1}^n a_i^2$ and $1 < OB^2 = \mathbf{v} \cdot \mathbf{v} = \sum_{i=1}^n b_i^2$, as desired.