No, it's not $a=b=c=d$

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

We can apply AM-GM directly:

$AM(a+d,b,c) \geq GM(a+d,b,c) \quad\Rightarrow \quad \dfrac{a+d+b+c}3 \geq \sqrt[3]{bc(a+d)} \quad\Rightarrow \quad \dfrac{15}3 \geq \sqrt[3]{abc+bcd} \quad\Rightarrow \quad abc + bcd \leq 5^3 = \boxed{125}$

Equality holds when $b=c=a+d=5$ .

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

We can rewrite our given as $bc(a+d)$

By AM-GM,

$\large \frac{b+c}{2} \geq \sqrt{bc}$ , so $b+c \geq 2\sqrt{bc}$

Adding $a+d$ to both sides gives us

$a+b+c+d \geq 2\sqrt{bc}+(a+d)= \sqrt{bc}+\sqrt{bc}+(a+d)$

By AM-GM again,

$\large \frac{\sqrt{bc}+\sqrt{bc}+(a+d)}{3} \geq \sqrt[3]{\sqrt{bc} \cdot \sqrt{bc} \cdot (a+d)}= bc(a+d)$

Multiplying 3 by both sides gives us

$2\sqrt{bc}+(a+d) \geq 3\sqrt[3]{bc(a+d)}$

But, since $a+b+c+d \geq 2\sqrt{bc}+(a+d)$ , then $a+b+c+d \geq 3\sqrt[3]{bc(a+d)}$

So, $5=\frac{15}{3}= \frac{a+b+c+d}{3} \geq \sqrt[3]{bc(a+d)}$

Cubing both sides gives us $125 \geq bc(a+d)$

So, our answer is $\boxed{125}$

Define $x = a + d$ , then we must maximize $xbc$ under the condition $x + b + c = 15$ . The solution is well know: $x = b = c = 5$ and $xbc = \boxed{125}$ .

$a+d=x$

$abc+bcd = bcx$

$x+b+c = 15$

$\frac{x+b+c}{3} \geq \sqrt[3]{bcx}$

$5^3 = 125 \geq bcx$

$EZ$