Question:
For $a,b,c,d > 0$ and $abcd = 1$, show that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$$
My Attempts:
Trivial to see that equality occurs for $a = b= c = d$ – so this indicates to use some inequality wherin the equality condition holds when all quantities are equal – I've tried AM-GM-HM, Power means, Rearrangement, Cauchy-Schwarz, Newton, Maclaurin, Weighted AM-GM – it doesn't help, and the reason is that these inequalities obtain a function of $a,b,c,d$ on the RHS whose minimum is $<7$ – and that minimum and the inequality can never hold simultaneously – giving a loose bound.
I've tried rearranging the inequality to the form $(\sum_{cyc} a )(\sum_{cyc} \frac{1}{a}) – 7 (\sum_{cyc} a) + 12 \geq 0 $ – This doesn't help. I've tried the substitution $(a,b,c,d) \rightarrow (\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{w} )$ – this is of no help either. Even tried the substitution $(a,b,c,d) \rightarrow (\frac{x}{y}, \frac{y}{z}, \frac{z}{w}, \frac{w}{x} )$ – of no help either. Writing it as $\frac{4}{M(-1)} + \frac{3}{M(1)} \geq 7$ where $M(x)$ is the power means function, and hoping to use some property of $M(x)$ doesn't work either.
My Analysis: Suppose $S_i$ denotes the sum taken $i$ at a time. It is trivial to see that $S_3 = \sum_{cyc} \frac{1}{a}$ is lower bounded by $S_4$ by Maclaurin or GM-HM. However $S_1$ is unbounded above even if $S_4 = 1$. However when $S_1$ reaches a very high value, so does $S_3$, so a minimum has to involve considering both terms simultaneously i.e. you can't minimize each separately and then hope to minimize the function.
Brute Force Solution: The brute force solution of considering it as a function of 4 variables and then optimizing obviously works, I'm hoping for something more smarter or elegant.
Best Answer
Mixing Variables and $uvw$ help.
I'll post my solution, which I found eight years ago.
Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$.
Thus, $$f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$$ $$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(a+b+c+d)\left(a+3\sqrt[3]{bcd}\right)}\geq$$ $$\geq\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(\frac{b+c+d}{3}+b+c+d)\left(\frac{b+c+d}{3}+3\sqrt[3]{bcd}\right)}=$$ $$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}.$$ We'll prove that $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0.$$ Indeed, let $b+c+d=3u$, $bc+bd+cd=3v^2$ and $bcd=w^3$.
Thus, $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0$$ it's $ g(v^2)\geq0,$ where $g$ is a linear increasing function.
Id est, $g$ gets a minimal value, when $v^2$ gets a minimal value, which happens for equality case of two variables.
Since $g(v^2)\geq0$ is homogeneous inequality, it's enough to check one case only: $c=d=1$, which after substitution $b=x^3$ gives $$(x-1)^2(2x^7+x^6+18x^5-10x^4-50x^3+36x^2+26x+4)\geq0,$$ which is true.
Id est, $$f(a,b,c,d)\geq f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)$$ and it's enough to prove that $f(a,b,b,b)\geq0$, where $a=\frac{1}{b^3}$, which gives $$(b-1)^2(3b^6+6b^5+9b^4-9b^3-5b^2-b+3)\geq0,$$ which is true.