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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.

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    $\begingroup$ you should show your efforts first, and you've been here long enough to know the way things are done.... $\endgroup$ – DeepSea Jul 11 '15 at 5:41
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hint: Use the AM-GM inequality: $4+a=1+1+1+1+a\geq 5\sqrt[5]{a}$

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  • $\begingroup$ Yes, Ans. is 256, but for what condition when minimium value holds. $\endgroup$ – Hardey Pandya Jul 11 '15 at 5:44
  • $\begingroup$ when $a=b=c=d=4$, ans. is 4096. Which is very greater than 256. also abcd is not equal to 1. $\endgroup$ – Hardey Pandya Jul 11 '15 at 5:47
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    $\begingroup$ I fixed it. Check out my edit ! $\endgroup$ – DeepSea Jul 11 '15 at 5:50
  • $\begingroup$ Okay, thanks... $\endgroup$ – Hardey Pandya Jul 11 '15 at 5:52
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Use Holder's inequality $$(4+a)(4+b)(4+c)(4+d)\ge(4+\sqrt[4]{abcd})^4=625$$ Equality is when $a=b=c=d=1$.

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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 $a=b=c=d=1$.

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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 $t = 1$.

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