<s>[INST] If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$? What I know is that for equations of type $x+y=8$, $xy$ attains its maximum value when $x=y$ and this can be proved by either solving the quadratic equation with completing the squares or finding the first and second derivatives of $xy$ w.r.t. $x$ to perform maxima/minima test. But both of these methods are applicable only when there is not more than one independent variable -- here $x$ is independent variable and $y=8-x$ is the dependent one.
I don't understand why $(a+1)(b+1)(c+1)(d+1)$ attains its maximum value at $a=b=c=d$ within the condition of $a+b+c+d=1$. How can we prove this fact? In some other similar posts I saw others using the Lagrange multiplier method. This method is too advanced for me to understand. I'm looking for some kind of algebraic proof similar to comleting with sqaure proof as in case of finding the maximum value of $xy$.
 [/INST] You known AM-GM inequality?
$$(a+1)(b+1)(c+1)(d+1)\le\left(\dfrac{a+b+c+d+4}{4}\right)^4$$
if you don't known AM-GM inequality,you also can following
In fact ,if $x,y,z,w\in R^{+}$,then we have
$$x^4+y^4+z^4+w^4-4xyzw\ge 0$$
because
$$x^4+y^4+z^4+w^4-4xyzw=(x^2-y^2)^2+(z^2-w^2)+2(xy-zw)^2$$
so $=$ iff $x=y=z=w$
so
$$4xyzw\le \le x^4+y^4+z^4+w^4$$
let
$$\sqrt[4]{a+1}=x,\sqrt[4]{b+1}=y,\sqrt[4]{c+1}=z,\sqrt[4]{d+1}=w$$
so
$$4\sqrt[4]{(a+1)(b+1)(c+1)(d+1)}\le a+1+b+1+c+1+d+1=5$$
so
$$(a+1)(b+1)(c+1)(d+1)\le \left(\dfrac{5}{4}\right)^4$$
</s>