Since nobody else has posted, here is my attempt Let $X = (x^2 + y^2)$, and $A = (z^2 + t^2 )$ and $P = (xz + yt)$ We are told that $(X-1)(A-1)>(P-1)^2$. Also by CS, $(X+1)(A+1)>(P+1)^2$ Multiplying these two inequalities gives $(X^2-1)(A^2-1)>(P^2-1)^2$ (*) Clearly for (*) to be true, since the right hand side is positive, either $X>1$ and $A >1$ (in which case we are done), or $X<1$ and $A <1$ Let us assume the latter, that $X<1$ and $A <1$, and show a contradiction. By CS, $P^2<AX$. so let $P^2=AX-d$ where d is positive. Substitute for P into (*) gives $2AXd - d^2 -2d>(A-X)^2 >0$ Therefore $2AX>2+d$, therefore $AX>1$, which contradicts our original assumption that $X<1$ and $A <1$

Let $a,b,c,d$ be real numbers such that $a^2 + b^2 \le 1 $ and $c^2 + d^2 \le 1.$ Prove that $$(a^2 + b^2 -1)(c^2 +d^2 - 1) \leq(ad +bc -1)^2$$https://artofproblemsolving.com/community/c6h1164949p5561671 h h