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If $a,b,c,d$ are positive real numbers and $abcd=1$ then Find the minimum value of $(4+a)(4+b)(4+c)(4+d)$. Find the condition when minimium value holds.
I've used AM-GM Inequality $4+a \ge 2 \sqrt{4a}$, $4+b \ge 2 \sqrt{4b}$, $4+c \ge 2 \sqrt{4c}$, $4+d \ge 2 \sqrt{4d}$. Since $a,b,c,d$ are positive we can multiply these inequalities... $(4+a)(4+b)(4+c)(4+d) \ge 256$. I stuck at when finding the condition for equality holds.
hint: Use the AM-GM inequality: $4+a=1+1+1+1+a\geq 5\sqrt[5]{a}$
Use Holder's inequality $$(4+a)(4+b)(4+c)(4+d)\ge(4+\sqrt[4]{abcd})^4=625$$ Equality is when $a=b=c=d=1$.
The answer is not in fact 256; you can't actually attain it. You can find the stronger bound 625 by using the fact that by weighted AM-GM, $\frac{1}{5}a + \frac{4}{5} \cdot 1 \ge a^{1/5}$, so $a+4 \ge 5a^{1/5}$
\begin{align*} \prod_\text{cyc}(a+4) &\ge \prod_\text{cyc}5a^{1/5} \\ &= 625(abcd)^{1/5} \\ &= 625 \end{align*}
Thus the given product is actually always at least 625, so you can't get 256. You can get 625 by setting $a=b=c=d=1$.
The function $f(x_1, \dots, x_4) = - (4 + x_1) \cdots (4 + x_4)$ is convex on the convex space $S = \{x_1, \dots, x_4\geq 0:\, x_1 \cdots x_4 \geq 1\}$. Clearly $f\to -\infty$ as $\min |x_i| \to \infty$. Since $f$ isn't constant on any lines, it must therefore have a single global maximum $p\in S$. But $f$ is symmetric under permuting the $x_i$, so $p$ must be invariant under all those permutations. Hence $p = (t, \dots, t)$ for some $t \geq 1$. Since $f(t, \dots, t)$ is decreasing in $t$, we have $t = 1$.
