$(1-a)(1-b)(1-c)(1-d)\geq abcd$ for $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that $$(1-a)(1-b)(1-c)(1-d)\geq abcd.$$
I thought about substituting $a=\sqrt{w},b=\sqrt{x}$, etc. (assuming first that $a,b,c,d$ are positive), and then looking at the convexity of the function $f(r)=(1-\sqrt{r})/\sqrt{r}$ and applying some Jensen-type inequality. But such an inequality applies to the sum of functions, not the product.

Observe that  as long as $a,b,c,d \ge 0$: $$\begin{align}(1-a)(1-b) - cd &\ge 1 - a - b + ab - \left(\frac{c^2+d^2}{2}\right) \\ &= 1 - a - b + ab - \left(\frac{1-a^2-b^2}{2}\right) \\ &= \frac{1}{2}(1-a-b)^2 \ge 0\end{align}$$
Similarly, $$(1-c)(1-d) -ab \ge \frac{1}{2}(1-c-d)^2\ge 0$$
Hence, $$(1-a)(1-b)(1-c)(1-d) \ge abcd$$
