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: 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.
A: $a^2+b^2+c^2+d^2=1$ => $a,b,c,d\in[0,1]$.
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
=>$xyzw\geq1$
A: Using AM-GM we can write $1-a^2-b^2=c^2+d^2\ge2cd$
So the idea right is to prove this $(1-a)(1-b)\ge cd$
or similary Proving this $2(1-a)(1-b) -2cd\ge 0$
$$2(1-a)(1-b) -2cd \ge 2(1-a)(1-b) -1 +a^2+b^2 = (1-a-b)^2\ge 0$$
So $(1-a)(1-b)\ge cd$ is true.Similarly we can show $(1-c)(1-d)\ge ab$. multiply them to obtain the question inequality.
