# About an inequality including arithmetic mean, geometric mean and harmonic mean

For any $n$ positive real numbers $a_i\ (i=1,2,\cdots,n)$, let us define $A,G,H$ as $$A=\frac{\sum_{i=1}^{n}a_i}{n},\ G=\sqrt[n]{\prod_{i=1}^{n}a_i},\ H=\frac{n}{\sum_{i=1}^{n}\frac{1}{a_i}}.$$

Then, here is my question.

Question : How can we find the minimum value of a real number $p$ which satisfies the following inequality for any $a_i(i=1,2,\cdots,n)$? $$pA+(1-p)H\ge G$$

I've already found that the answer for $n=2$ is $p=1/2$. However, I haven't had any good idea for $n\ge 3$ in general. Can anyone help?

• For $n=2$ we have the special case $G=\sqrt{AH}$. Dividing by H and letting $x=\sqrt{A/H}$ we have to solve a simple quadratic $($in$)$equation. May 22 '14 at 2:59
• Conjecture: $p_n = 1-\dfrac1n$. Seems good for $n=2, 3, \infty$, though need to check more. May 22 '14 at 8:08
• You may also want to check math.stackexchange.com/questions/92935/am-gm-hm-triplets May 22 '14 at 9:44
• In this answer math.stackexchange.com/a/176909/448 I sketch a way to look at this, and argue that it may be hard. May 25 '14 at 11:15

By Maclaurin's Inequality, $S_1 \ge \sqrt[n-1]{S_{n-1}} \implies S_1^{n-1} \ge S_{n-1} \implies A^{n-1}H \ge G^n$.

So if we normalise with $A=1$, we know $0 < H \le G \le A = 1$ and we have $H \ge G^n$ with equality when $H=G=A=1$.

The normalised inequality is $p+(1-p)H \ge G$, and with the above bound, it is enough to have $p+(1-p)G^n \ge G$. As this must hold $\forall G \in (0, 1]$, we must have: $$p \ge \frac{G-G^n}{1-G^n}$$

As the RHS achieves a maximum of $1-\frac1n$ when $G \to 1$, we need $p_n \ge 1-\frac1n$. Thus we have:$$\left(1-\tfrac1n \right)A+\tfrac1nH\ge G$$

Updated based on David Speyer's comment and post, as an upper bound rather than the optimal $p_n$ ...

For $n=3, 4, 5, ...10$, the minimal $p_n$ seem to be $\approx$ \begin{array}{ l | c } \hline 3 & 0.52605 \\ \hline 4 & 0.56301 \\ \hline 5 & 0.59660 \\ \hline 6 & 0.62560 \\ \hline 7 & 0.65055 \\ \hline 8 & 0.67213 \\ \hline 9 & 0.69095 \\ \hline 10 & 0.70752 \\ \hline \hline \end{array}

P.S. The table above was calculated by numerically finding the maximum of $$p_n = \max_{x>1} \frac{G_n-H_n}{1-H_n} = \max_{x>1} \frac{\sqrt[n]{x^{n-1}(n-(n-1)x)}-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}{1-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}$$

• A nice conjecture. Thanks. May 23 '14 at 11:42
• A very nice answer! It isn't optimal though even for $n=3$, (unless I screwed up), see my answer. May 25 '14 at 11:43
• I think that this is optimal (except in the case of $n=5$, where I think there is something amiss). See my answer.
– robjohn
May 30 '14 at 0:59
• @DavidSpeyer: try two samples at $1.25195252$ and one at $0.49609496$. I believe we get $p=0.52604989$. I generated these values using this answer.
– robjohn
May 30 '14 at 2:10
• @robjohn I am saying that $1-1/n$ isn't optimal. It sounds like you agree: You get $0.526$, not $1-1/3=0.666$. May 30 '14 at 3:56

To demonstrate that finding the optimal bound is hard, I'll work out the case $n=3$. We already found out that the optimum will occur at a point of the form $(x,x,y)$ with $x>y$. I'll normalize $G=1$, so $y=1/x^2$. Then $$A = \frac{2x + x^{-2}}{3} \quad H = \frac{3}{2x^{-1} + x^2}.$$ We want to optimize $$\theta = \frac{1-H}{A-H} = \frac{3 x^2 (2+x)}{2(1+x+x^2)^2}.$$ Macavity's bound, which is very slick and awesome, is to replace $H$ by $G^3/A^2 = 1/A^2$ and consider $\phi = \frac{1-1/A^2}{A-1/A^2}$ instead.

Here is a plot of both $\theta$ and $\phi$ for $1 \leq x \leq 3$ ($\theta$ is blue, $\phi$ is red). Note that $\phi > \theta$, $\phi$ is decreasing and $\phi(1) = 2/3$, as Macavity computes. However, $\max \theta < 2/3$.

We have $$\frac{d \theta}{dx} = \frac{3 x (4 + 3 x - 3 x^2 - x^3)}{2 (1 + x + x^2)^3}$$ The global maximum is at the root of $4+3x-3x^2-x^3$ which is roughly $1.36147$. The corresponding optimum for $\theta$ is roughly $0.52605$. (Of course, you can get closed form answers using the cubic formula, but they aren't enlightening.)

I still think that finding nice functions $p_n$ which work is a good project, but I don't think you'll get a closed form for the optimal answer.

Some more thoughts: Repeating this computation with $n$ in place of $3$, we have $$\frac{d \theta}{d x} = \frac{ n x^{n-2} \cdot f(x)}{(n-1) (1-x^n)^3}$$ where $$f(x) := (n^2-2n+1) - n^2 x + (n^2+2n-2) x^n - n^2 x^{n+1}.$$ By Descartes' Rule of signs, $f$ only has four positive roots. We can compute that $f(x)$ has a triple root at $x=1$, which cancels the triple root of $(1-x^n)^3$, so there is only one positive root to care about.

I set $r=x/y$, since this is independent of our choice of normalization (so $r=x^n$ if we normalize $G=1$.) Computations for $3 \leq n \leq 100$ suggest that $r$ grows something like $n \log n$. I've copy pasted the numbers below in the format $\{ n, r \}$ case someone wants to play with them:

{{3, 2.52361}, {4, 4.34916}, {5, 6.40716}, {6, 8.65639}, {7, 11.0686}, {8, 13.6232}, {9, 16.3041}, {10, 19.0987}, {11, 21.9967}, {12, 24.9895}, {13, 28.0699}, {14, 31.2317}, {15, 34.4694}, {16, 37.7785}, {17, 41.1546}, {18, 44.5942}, {19, 48.094}, {20, 51.6509}, {21, 55.2622}, {22, 58.9256}, {23, 62.6388}, {24, 66.3997}, {25, 70.2064}, {26, 74.0572}, {27, 77.9504}, {28, 81.8847}, {29, 85.8585}, {30, 89.8706}, {31, 93.9197}, {32, 98.0047}, {33, 102.124}, {34, 106.278}, {35, 110.464}, {36, 114.683}, {37, 118.932}, {38, 123.212}, {39, 127.521}, {40, 131.859}, {41, 136.225}, {42, 140.618}, {43, 145.039}, {44, 149.485}, {45, 153.957}, {46, 158.454}, {47, 162.975}, {48, 167.521}, {49, 172.09}, {50, 176.681}, {51, 181.296}, {52, 185.933}, {53, 190.591}, {54, 195.271}, {55, 199.971}, {56, 204.693}, {57, 209.434}, {58, 214.196}, {59, 218.976}, {60, 223.776}, {61, 228.595}, {62, 233.433}, {63, 238.289}, {64, 243.163}, {65, 248.054}, {66, 252.963}, {67, 257.889}, {68, 262.833}, {69, 267.792}, {70, 272.769}, {71, 277.761}, {72, 282.77}, {73, 287.794}, {74, 292.834}, {75, 297.89}, {76, 302.96}, {77, 308.046}, {78, 313.146}, {79, 318.261}, {80, 323.39}, {81, 328.533}, {82, 333.691}, {83, 338.862}, {84, 344.047}, {85, 349.246}, {86, 354.458}, {87, 359.683}, {88, 364.922}, {89, 370.173}, {90, 375.437}, {91, 380.714}, {92, 386.003}, {93, 391.305}, {94, 396.619}, {95, 401.945}, {96, 407.283}, {97, 412.634}, {98, 417.995}, {99, 423.369}, {100, 428.754}}


If $r \approx n \log n$ is correct, and I choose $(1,1,1,\ldots, 1, 1/(n \log n))$ as my representative, then I get $$A \approx 1 - 1/n, \ G \approx 1-\log n /n,\ \ H \approx 1/\log n \ \mbox{and} \ \theta \approx 1-\log n/n.$$ But finding explicit inequalities that achieve this behavior doesn't sound fun to me.

In this answer, it is shown that for $n$ items with an arithmetic mean of $1$ and a harmonic mean of $h$, the geometric mean is between $g(h,\frac1n)$ and $g(h,\frac{n-1}n)$ where $$g(h,\lambda)=\frac{\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+1\right)^\lambda\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}-1\right)^{1-\lambda}}{\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+2\lambda-1}\tag{1}$$ We can compute the minimum $p$ that will work for a given $n$ and $h$ with $$p(n,h)=\frac{g(h,\frac{n-1}{n})-h}{1-h}\tag{2}$$ Maximizing $p(n,h)$ for $h\in[0,1]$, we get $$\begin{array}{r|c|c} n&p&h\\\hline 2&0.50000000&1.00000000\\\hline 3&0.52604991&0.83027796\\\hline 4&0.56301305&0.67404567\\\hline 5&0.59659653&0.57799531\\\hline 6&0.62560358&0.51531972\\\hline 7&0.65054843&0.47136304\\\hline 8&0.67212783&0.43872876\\\hline 9&0.69095251&0.41342714\\\hline 10&0.70751470&0.39314442\\\hline \end{array}\tag{3}$$ which agrees with Macavity's answer.