# For $a,b,c,d > 0$ and $abcd = 1$, show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$

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.

• Great job telling us what you've already tried in the question statement! – Greg Martin Jun 5 '19 at 18:14

## 1 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.

• I'll take me a while to go through this - but could you say give some insight as to how you formed your approach - was it a lucky idea or is there some theory behind it (cause I'm still a beginner, I wouldn't know) - i.e what ideas could I take back from this question? No offence it's a great method, but tbh optimizing over 4 variables would be more concise. – Kaind Jun 5 '19 at 18:37
• @Kaind It was optimizing over 4 variables. There are very many methods of inequalities proofs, but non of those methods are working in the general. In our case the Mixing Variables method helps. – Michael Rozenberg Jun 5 '19 at 18:43
• There's something called a 'mixing variables method' - could you point me to some link or pdf which explains this? (By optimizing 4 variables, I meant finding saddle points and critical points, and then getting answer). – Kaind Jun 5 '19 at 18:45
• For those questions visit the AOPS website. – Dr. Sonnhard Graubner Jun 5 '19 at 18:52
• @Kaind About MV you can read here: math.stackexchange.com/questions/tagged/mixing-variables Now I see that Lagrange multipliers helps, but for the proof we need to consider two cases: 1. $b=c=a$; 2. $b=a$ and $c=d$. – Michael Rozenberg Jun 5 '19 at 18:52