If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$

Question

If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$

Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that this is true, but I don't seem to be able to prove it.

Answer 1

Hint: If $x$, $y\le1$ then $$0\le(1-x)(1-y)=1-x-y+xy$$

Answer 2

Let $x = a^2+b^2, y = c^2+d^2 \implies x+y \le 1+xy \implies (x-1)(1-y) \le 0$, which is true since $ 0 \le x,y \le 1$.