## A5.

Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.

Find the minimum value and the maximum value of the product $a b c d$.


Solution. Let's find the minimum first.

$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$

By AM-GM, $4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1$.

Note that if $a=b=c=1$ and $d=-1$, then $a b c d=-1$.

We'll find the maximum. We search for $a b c d>0$.

Obviously, the numbers $a, b, c$ and $d$ can not be all positive or all negative.

WLOG $a, b>0$ and $c, d0, a+b-x-y=2$ and $a^{2}+b^{2}+x^{2}+y^{2}=4$. We need to find $\max (a b x y)$. We get: $x+y=a+b-2$ and $x^{2}+y^{2}=4-\left(a^{2}+b^{2}\right)$. Since $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)$, then $2\left(a^{2}+b^{2}\right)+(a+b-2)^{2} \leq 8$; on the other hand, $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right) \Rightarrow(a+b)^{2}+(a+b-2)^{2} \leq 8$.

Let $a+b=2 s \Rightarrow 2 s^{2}-2 s-1 \leq 0 \Rightarrow s \leq \frac{\sqrt{3}+1}{2}=k$.

But $a b \leq s^{2} \Rightarrow a b \leq k^{2}$.

Now $a+b=x+y+2$ and $a^{2}+b^{2}=4-\left(x^{2}+y^{2}\right)$. Since $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$, then $2\left(x^{2}+y^{2}\right)+(x+y+2)^{2} \leq 8$; on the other hand, $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right) \Rightarrow$

$\Rightarrow(x+y)^{2}+(x+y+2)^{2} \leq 8$. Let $x+y=2 q \Rightarrow 2 q^{2}+2 q-1 \leq 0 \Rightarrow q \leq \frac{\sqrt{3}-1}{2}=\frac{1}{2 k}$.

But $x y \leq q^{2} \Rightarrow x y \leq \frac{1}{4 k^{2}}$.

In conclusion, abxy $\leq k^{2} \cdot \frac{1}{4 k^{2}}=\frac{1}{4} \Rightarrow a b c d \leq \frac{1}{4}$.

Note that if $a=b=k$ and $c=d=-\frac{1}{2 k}$, then $a b c d=\frac{1}{4}$

In conclusion, $\min (a b c d)=-1$ and $\max (a b c d)=\frac{1}{4}$.

