# Conditions under which inequality holds for all triplets of non-negative reals

I have the following inequality:

$$\left(\frac{1}{3} \left( \left(\frac{a+b}{2}\right)^3 + \left(\frac{a+c}{2}\right)^3 + \left(\frac{b+c}{2}\right)^3 \right) \right)^\frac{1}{3} \leq \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p}$$ which is true $\forall a,b,c \in \mathbb{R}^+\cup\{0\}$.

I would like to show that this inequality above holds if and only if $p \geq \frac{3}{2}$.

i.e. the inequality holds for all non-negative reals $a,b,c$ as long as $p \geq \frac{3}{2}$.

Any help whatsoever would be appreciated, so please comment even if you don't have a full solution (even proving 1 direction of the implication would be very helpful).

P.S. I'm not sure exactly what to tag this, I've tagged it as convex-optimisation for now, as one of the things I've tried unsuccessfully is maximising/minimising one the sides subject to the constraint of the other side being held constant. If this is tagged wrong, please let me know the right tag(s).

• Which implication is the one of most interest to you? Also, could you perhaps say something about how this conjectured inequality arose? Jul 15, 2017 at 2:55
• The inequality implying that p is at least 3/2 is of slightly more interest. The inequality came about in a paper (not by me) about the L^p-improving properties of the convolution operator generated by the Cantor Measure. The inequality holds for some value of p. We got the specific value of p using numerical methods. We are very certain we have the correct values for p, but we are having trouble proving it. Jul 15, 2017 at 3:11
• The inequality is trivial for $p\ge 3$ (just Holder and triangle inequalities). Jul 15, 2017 at 4:58
• We can rephrase the question as: For what $p$ is $\max (a+b)^3+(b+c)^3+(c+a)^3\le 24$, where the maximization is under the constraint $a^p+b^p+c^p=3$? I'm not sure one can do a whole lot with this, other than investigate it numerically, as you did. Jul 15, 2017 at 5:59
• @ChristianRemling: That maximisation is basically exactly what we did to find the $p \geq \frac{3}{2}$. We would like to prove it, as it would be quite a nice result, but unfortunately it's proven quite difficult. Jul 15, 2017 at 7:05

it would be quite a nice result, but unfortunately it's proven quite difficult.

I'm not sure about "nice" (after all, one can invent infinitely many inequalities for 3 positive numbers) but it is certainly not difficult.

WLOG, $a+b+c=3$. Write $a=1+A, b=1+B, c=1+C$ with $A,B,C\ge -1, A+B+C=0$. Then the LHS equals $$\left[\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3\right]^{1/3}$$

Part 1: $p$ must be $\ge 3/2$

Consider the case when $A,B,C\to 0$ and look at both sides up to the second order. We have $$LHS=1+\frac{A^2+B^2+C^2}{12}+\text{higher order terms}\,; \\ RHS=1+\frac{p-1}6(A^2+B^2+C^2)+\text{higher order terms}$$ whence $p-1\ge \frac 12$ is necessary.

Part 2: $p=3/2$ works (and, thereby, any larger $p$ works as well by Holder).

We need to prove that $$\sqrt{\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3} \le \frac{(1+A)^{3/2}+(1+B)^{3/2}+(1+C)^{3/2}}3$$
Opening the parentheses, recalling that $A+B+C=0$, and using the inequality $\sqrt{1+X}\le 1+\frac X2$, we estimate the LHS from above by $$1+\frac{A^2+B^2+C^2}8-\frac{A^3+B^3+C^3}{48}$$ Thus, it would suffice to show that $$(1+A)^{3/2}\ge 1+\frac 32A+\frac 38A^2-\frac 1{16}A^3$$ However the RHS here is just the cubic Taylor polynomial of $A\mapsto (1+A)^{3/2}$ at $A=0$ and the fourth derivative is positive.

• I'm not sure about "nice" (after all, one can invent infinitely many inequalities for 3 positive numbers) Yes, of course one can come up with as many random inequalities as one wants. I should have said "Within the context of a larger problem I'm working on, proving this will give us a nice result." Thanks for the solution by the way, very clever. Jul 16, 2017 at 16:55
• @CoffeeTableEspresso You are welcome :-). The solution is fairly standard, by the way, especially part 1 (what can be more natural than analyzing the second differential at the point of the suspected maximum?). The trickery in the second part was just to write the inequality in the form that allows one to make simple estimates that are still exact up to the third order (since the quadratic terms coincide, you cannot afford any sloppiness with the third order terms, so your next hope is the fourth order). This was just to explain how I did it, so that you'll be able to do the next one yourself. Jul 16, 2017 at 17:37
{"ft_lang_label":"__label__en","ft_lang_prob":0.86905605,"math_prob":0.99920136,"size":952,"snap":"2022-27-2022-33","text_gpt3_token_len":293,"char_repetition_ratio":0.13607594,"word_repetition_ratio":0.0,"special_character_ratio":0.31407562,"punctuation_ratio":0.09268293,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99973434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T17:13:20Z\",\"WARC-Record-ID\":\"<urn:uuid:acff56a5-5b32-49ca-9d64-e8ac5a80c940>\",\"Content-Length\":\"111208\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:734b7c38-c4f1-41b5-9367-e9982448b91a>\",\"WARC-Concurrent-To\":\"<urn:uuid:70a578c5-cc29-4aea-9076-5c44ce117b44>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/275468/conditions-under-which-inequality-holds-for-all-triplets-of-non-negative-reals\",\"WARC-Payload-Digest\":\"sha1:WMVGI4MMJ7LVSKJDX6H4JZRFTRWWQDGS\",\"WARC-Block-Digest\":\"sha1:M44KHGGBDUYUAYKF425AXGB7EG6HTPNE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571056.58_warc_CC-MAIN-20220809155137-20220809185137-00001.warc.gz\"}"}