15
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lucian May 22 '14 at 2:59
  • $\begingroup$ Conjecture: $p_n = 1-\dfrac1n$. Seems good for $n=2, 3, \infty$, though need to check more. $\endgroup$ – Macavity May 22 '14 at 8:08
  • 2
    $\begingroup$ You may also want to check math.stackexchange.com/questions/92935/am-gm-hm-triplets $\endgroup$ – Macavity May 22 '14 at 9:44
  • 1
    $\begingroup$ In this answer math.stackexchange.com/a/176909/448 I sketch a way to look at this, and argue that it may be hard. $\endgroup$ – David E Speyer May 25 '14 at 11:15
7
$\begingroup$

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 $$p_n = \max_{x>1} \frac{G_n-H_n}{1-H_n} = \max_{x>1} \frac{\sqrt[n]{x^{n-1}(n-(n-1)x)}-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}{1-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}$$

$\endgroup$
  • $\begingroup$ A nice conjecture. Thanks. $\endgroup$ – mathlove May 23 '14 at 11:42
  • $\begingroup$ A very nice answer! It isn't optimal though even for $n=3$, (unless I screwed up), see my answer. $\endgroup$ – David E Speyer May 25 '14 at 11:43
  • $\begingroup$ I think that this is optimal (except in the case of $n=5$, where I think there is something amiss). See my answer. $\endgroup$ – robjohn May 30 '14 at 0:59
  • $\begingroup$ @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. $\endgroup$ – robjohn May 30 '14 at 2:10
  • $\begingroup$ @robjohn I am saying that $1-1/n$ isn't optimal. It sounds like you agree: You get $0.526$, not $1-1/3=0.666$. $\endgroup$ – David E Speyer May 30 '14 at 3:56
9
+25
$\begingroup$

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, $\max \theta < 2/3$.

enter image description here

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.