Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$
Prove
$\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$
I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work.

Observe that $f(x)=\log\left(\frac1x-1\right)$ is concave for $0<x<1$ (its derivative is $\frac1{x(1-x)}$).
Taking $\log$, the inequality is equivalent to
$$\sum\log\left(\frac1a-1\right)\geq0.$$
By Jensen's inequality,
$$\sum\log\left(\frac1a-1\right)\geq4\log\left(\frac4{a+b+c+d}-1\right).$$
By Cauchy-Schwarz/Power Mean/Hölder, $a+b+c+d\leq2\sqrt{a^2+b^2+c^2+d^2}=2$, hence
$$LHS\geq4\log\left(\frac42-1\right)=0,$$
as was to be shown.
