By AM-GM, $\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \dfrac{(1-a)(1-c)+(1-b)(1-d)}{2}=\dfrac{ac+bd+2-(a+b+c+d)}{2}=\dfrac{ac+bd}{2},$ as desired.

1st solution: $\sqrt{(1-a)(1-b)(1-c)(1-d)}\overset{\text{AM-GM}}{\le}\frac{(1-b)(1-d)+(1-a)(1-c)}{2}=\frac{ac+bd+2-\sum_{cyc}a}{2}=\frac{ac+bd}{2}$ $\blacksquare$ 2nd solution: $\sqrt{(1-a)(1-b)(1-c)(1-d)}\overset{\text{AM-GM}}{\le}\frac{ab+dc}{2}$ thus the inequality is equivalent to: $ab+cd\le ac+bc\Longleftrightarrow (a-d)(c-b)\ge0$ Furthermore let: $f(a,b,c,d)=ac+bd-ab-cd=(a-d)(c-b), g(a,b,c,d)=a+b+c+d-2, \text{and}\Rightarrow L=f(a,b,c,d)-\lambda g(a,b,c,d)$ $$\frac{\partial L}{\partial a}=c-b-\lambda=0$$$$\frac{\partial L}{\partial b}=d-a-\lambda=0$$$$\frac{\partial L}{\partial c}=a-d-\lambda=0$$$$\frac{\partial L}{\partial d}=b-c-\lambda=0$$Thus $c-b=d-a=a-d=b-c$ (1) (2) (3) (4) respectively. From (2) and (3) we have that $a=d$ From (1) and (4) we have that $c=b$ Thus minimum is achieved when $f(a,c,c,a)=(a-a)(c-c)=ac+cd-ac-cd=0$ $\blacksquare$

Let $a, b, c, d\in(0,1)$ and $a+b+c+d=2$. Prove that$$\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{(a+b)(c+d)}{4} $$$$\sqrt{(1-a^2)(1-b^2)(1-c^2)(1-d^2)}\leq \frac{9(a+b)(c+d)}{16} $$

$\color{blue} \boxed{\textbf{SOLUTION}}$ By $\textbf{AM-GM Inequality,}$ $$\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{(1-a)(1-c)+(1-b)(1-d)}{2}=\frac{ac+bd+2-(a+b+c+d)}{2}=\dfrac{ac+bd}{2} \blacksquare$$

i hate ineq anyway, by AM-GM, $\frac{(1-a)(1-c)+(1-b)(1-d)}{2} \ge \sqrt{(1-a)(1-b)(1-c)(1-d)}$, follows, now all that's left is to prove, that $\frac{(1-a)(1-c)+(1-b)(1-d)}{2}=\frac{ac+bd}{2}$, note that $2-a-b-c-d+ac+bd=ac+bd$, using the substitution, $a+b+c+d=2$, and we are done.

EpicBird08 wrote:

Yes I was thinking the same thing! This a little too easy, no? I usuallly don't go on HSO

There seems to be a geometric interpretation a_507_bc wrote: Let $a$, $b$, $c$, $d$ be positive reals strictly smaller than $1$, such that $a+b+c+d=2$. Prove that \[ \sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{ac+bd}{2}. \] Build a cyclic quad $WXYZ$ with those side lengths. The left-hand side is the area by Brahmagupta formula; the right-hand side is $WY \cdot XZ / 2$ by Ptolemy theorem. So the quotient of LHS/RHS is $\sin(\angle(WY,XZ)) \leq 1$ as needed.

there is also a nice solution using substitution: $a=sin^2{x}, b=sin^2{y}, c=sin^2{z}, d=sin^2{t}$.

a_507_bc wrote: Let $a, b, c, d$ be positive reals strictly smaller than $1$, such that $a+b+c+d=2$. Prove that $$\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{ac+bd}{2}. $$ Bruh direct AM GM. Note that $(1-a)(1-c)+(1-b)(1-d)=ac+bd-a-b-c-d+2$