<s>[INST] 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.
 [/INST] 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.
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