inequalities - A plausible inequality.

The following proof is a bit heavy-handed; I'm sure you it can be simplified. Assume $a_1=1, a_n=0$ as suggested above and write:

Write

$$F(x,y) = sum_{i=1}^n a_i(x_i^2-y_i^2)$$ , $$G(x,y)=F(x,y)^2+langle x,yrangle^2.$$
Let $(x,y)in S^{n-1}times S^{n-1}$ be a point where $G$ is maximized, where we may assume that for each $i$ at least one of $x_i,y_i$ is non-zero. It is clear that $-sum_i y_i^2 leq F(x,y) leq sum_i x_i^2$ so we may also assume $langle x,yrangle neq 0$.

By the method of Lagrange multipliers there exist $xi,eta$ such that for all $i$

$$4a_i x_i F+2langle x,yrangle y_i=2xi x_i$$ and
$$-4a_i y_i F+2langle x,yrangle x_i=2eta y_i.$$

Multiplying the first equation by $x_i$, the second by $y_i$, adding the two and summing over $i$ gives
$4G = 2(xi+eta)$. Multiplying the second equation by $y_i$, the first by $x_i$ and adding gives

$$langle x,yrangle (y_i^2+x_i^2) = (xi+eta)x_i y_i = 2Gcdot x_i y_i.$$ By assumption one of $x_i,y_i$ is non-zero. Dividing by the square of that number we see that the quadratic $$langle x,yrangle t^2 - 2G t + langle x,yrangle = 0$$ has a real root. Evaluating the discriminant it follows that $$G^2 leq langle x,yrangle^2 leq 1.$$

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