Minimum value of an expression

Question

What would be the minimum value of $$ (1+a)(1+b)(1+c)(1+d) \left({\frac{1}{a}}+{\frac{1}{b}}+{\frac{1}{c}}+{\frac{1}{d}}\right) $$ Given that $a,b,c,d$ are all positive real non zero number

THE ANSWER IS ${\frac{4^5}{27}}$

Answer 1

With symmetry, you expect $a=b=c=d$. So minimizing $f(a)=(1+a)^4(4/a)$ is quite straightforward.

Answer 2

By Symmetry an optimal value will be attained when $a=b=c=d$ so it suffice to consider \begin{eqnarray*} S=(1+a)^4\frac{4}{a}. \end{eqnarray*} Differentiation gives \begin{eqnarray*} \frac{dS}{da}=\frac{-4(1+a)^4}{a^2}+\frac{16(1+a)^3}{a}=\frac{4(1+a)^3(3a-1)}{a^2} \end{eqnarray*} Setting $\frac{dS}{da}=0$ gives $a=\frac{1}{3}$ and the minimum value is $\color{red}{\frac{4^5}{27}}$. (as stated in the question).

EDIT ... In order to show that an optimal value is attained when $a=b=c=d$. Calculate the partial derivatives \begin{eqnarray*} S&=&(1+a)(1+b)(1+c)(1+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\\ \frac{\partial S}{\partial a}&=&(1+b)(1+c)(1+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)+(1+a)(1+b)(1+c)(1+d)\frac{-1}{a^2} \end{eqnarray*} So we have \begin{eqnarray*} \frac{1+a}{a^2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1+b}{b^2}=\frac{1+c}{c^2}=\frac{1+d}{d^2} \end{eqnarray*} Now note that the function $f(x)=\frac{1+x}{x^2}$ is monotonic ... $f(x)=f(y)$ implies $x=y$ so $\color{blue}{a=b=c=d}$.