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.

$a^2+b^2+c^2+d^2=1$ => $a,b,c,d\in[0,1]$.
If $\displaystyle abcd=0$ so the inequality is true.
If $\displaystyle abcd>0$, set :
$x=\frac{1-a}{a},
y=\frac{1-b}{b},
z=\frac{1-c}{c},
w=\frac{1-d}{d}$
We have
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$
And the inequality become
$\displaystyle xyzw\geq1$
We will prove that with $x,y,z,w\geq0$ and $xyzw=1$:
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq1$
(1)
Using Cauchy-Schwarz we have
$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}\geq
\frac{1}{(\frac{x}{y}+1)(xy+1)}+\frac{1}{(\frac{y}{x}+1)(xy+1)}
+\frac{1}{(\frac{z}{w}+1)(zw+1)}+\frac{1}{(\frac{w}{z}+1)(zw+1)}
=\frac{1}{xy+1}+\frac{1}{zw+1}
=\frac{1}{xy+1}+\frac{1}{\frac{1}{xy}+1}
=\frac{1}{xy+1}+\frac{xy}{xy+1}=1$
Suppose that $xyzw<1$. Set $t=\frac{1}{xyz}$ so $xyzt=1$ and $t>w$. Using (1), we have
$1\le\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2}
<\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+w)^2}=1$
So the suppose is false
=>$xyzw\geq1$
