If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?

Question

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 $xy$.

Answer 1

Note that the maximum of $(5/4)^4$ is only if the variables are required to be bigger than $-1$. E.g. $(-10)+(-10)+10+11 = 1$ and $(-9)(-9)(11)(12) > (5/4)^4$.

Assuming the $\geq -1$ condition, using different variables ($u=a+1,v=b+1,w=c+1,x=d+1$) the problem can also be written as: given $u,v,w,x \geq 0$, $u+v+w+x =5$, what is the maximum value of $uvwx$?

The symmetry of the problem suggests that the max should occur when $u=v=w=x$, so the condition $u+v+w+x = 5$ implies they should all be equal to $5/4$. Now how do we show that this is actually the max?

Obviously the max does not occur when any of them are equal to zero, so let's exclude those cases. Suppose that two of them were not equal, i.e. $u \neq v$. Let's show that in this case we can strictly increase the product by replacing $u$ and $v$ with their average, $\frac{u+v}{2}$.

$$ \frac{u+v}{2}\frac{u+v}{2}wx > uvwx $$ $$ \iff u^2+v^2 + 2uv > 4uv $$ $$ \iff u^2 + v^2 -2uv > 0 $$ $$ \iff (u-v)^2 > 0 $$ $$ \iff \text{True} $$ Note that $(u-v)^2$ will be strictly bigger than zero because $u\neq v$. Thus we can strictly do better by replacing $u,v$ with their average. Thus, in order for $uvwx$ to be maximized, every pair of the variables must be equal. But then they are all equal, i.e. $u=v=w=x=\frac{5}{4}$ and you have your result.

Answer 2

Using AM, GM inequity

$$\dfrac{\sum (a+1)}4\ge\sqrt[4]{\prod(a+1)}$$

Answer 3

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 $$(a+1)(b+1)(c+1)(d+1)\le \left(\dfrac{5}{4}\right)^4$$

Answer 4

By the AM-GM inequality,

$$\sqrt[4]{(a+1)(b+1)(c+1)(d+1)} \leq \frac{(a + 1) + (b + 1) + (c + 1) + (d + 1)}{4} = \frac{5}{4}$$

hence...

Answer 5

In light of all the AM-GM answers, here is a variational approach.

Suppose we know that $a+b+c+d=1$, then we also know that for any direction moved $(\delta a,\delta b,\delta c,\delta d)$ that maintains this condition $$ \begin{align} 0 &=\delta a+\delta b+\delta c+\delta d\\ &=(\delta a,\delta b,\delta c,\delta d)\cdot(1,1,1,1)\tag{1} \end{align} $$ In that same direction, the rate of change of $(a+1)(b+1)(c+1)(d+1)$ satisfies $$ \begin{align} \small\frac{\delta\left[(a+1)(b+1)(c+1)(d+1)\right]}{(a+1)(b+1)(c+1)(d+1)} &=\small\frac{\delta a}{a+1}+\frac{\delta b}{b+1}+\frac{\delta c}{c+1}+\frac{\delta d}{d+1}\\ &\small=(\delta a,\delta b,\delta c,\delta d)\cdot\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)\tag{2} \end{align} $$ Case 1: $\boldsymbol{a,b,c,d\gt0}$ (interior maximum)

Suppose we are at the point where $(a+1)(b+1)(c+1)(d+1)$ is maximum given that $a+b+c+d=1$. Suppose further that we can find a direction that is perpendicular to $(1,1,1,1)$, but is not perpendicular to $\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)$. Then, since $a,b,c,d\gt0$, by $(1)$, we can travel in that direction, and its opposite, and $a+b+c+d$ will stay constant. However, by $(2)$, moving in that direction, or its opposite, will cause $(a+1)(b+1)(c+1)(d+1)$ to increase. This contradicts the maximality we assumed.

Thus, if $a,b,c,d\gt0$, $\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)$ must be parallel to $(1,1,1,1)$. That is, $a=b=c=d=\frac14$.

Case 2: one or more of $\boldsymbol{a,b,c,d=0}$ (boundary maximum)

Using the same arguments as above, we get that the non-zero variables must be equal. So we must not only check $\left(\frac14,\frac14,\frac14,\frac14\right)$, but also $\left(0,\frac13,\frac13,\frac13\right)$, $\left(0,0,\frac12,\frac12\right)$, and $(0,0,0,1)$.

Since $$ \left(\frac54\right)^4\ge\left(\frac43\right)^3\ge\left(\frac32\right)^2\ge2^1 $$ we get that the maximum occurs at $\left(\frac14,\frac14,\frac14,\frac14\right)$.