Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$

Question

(P1) Let $a, b, c$ be real numbers. Prove that $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge 0.$

This inequality is verified by Mathematica. I am particularly interested in (simple) SOS solutions (also for (P2) and (P3) below), however, any comments and solutions are welcome. I also gave a SOS solution for (P2) below, but it is complicated.

Some relevant questions:

(P2) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a+b+c)^6}{27}.$

Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$

(P3) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a^2+b^2+c^2+ab+bc+ca)^3}{8}.$

Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$

Wolfram|Alpha can't put the best $constant$ , but Mathematica (I used discriminant to find this inequality https://math.stackexchange.com/q/3138409/685744). — , Jun 04 '20 at 04:54
A big problem around $(a,b,c)=(1.05,-0.62,-1.15)$ for which the left side is closed to $0.111...$. — Michael Rozenberg, Jun 04 '20 at 05:19
@GiangNguyễnĐặngThanh Thanks. How is this question related to that question? — River Li, Jun 04 '20 at 05:35
@MichaelRozenberg I think that we may prove something like your inequality $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge k (a+b+c)^6$ for some constant $k$. — River Li, Jun 04 '20 at 05:37
@River Li It seems that it's true for $k=\frac{31}{81}$, but I have no a proof.At least, now we can write this inequality in the SOS's form. — Michael Rozenberg, Jun 04 '20 at 06:00
@MichaelRozenberg Yes, you are right which can be verified by Mathematica. — River Li, Jun 04 '20 at 06:05
Sometimes, one may easily express it as $u^\mathsf{T}Q u$ where $Q$ is constant symmetric rational matrix and $u$ is a vector containing monomials in variables. If you can do it, what is the size of $Q$? — River Li, Jul 14 '20 at 10:28
@nguyenhuyen_ag It is amazing! How powerful is it? Did you try complicated SOS? You may try my another question: https://math.stackexchange.com/questions/3711967/sum-of-squares-sos-solutions-for-a2b2c2d2abcd1-ge-abbccdda-acb — River Li, Jul 21 '20 at 02:46
@nguyenhuyen_ag By the way, for P2, I have posted one SOS solution: https://math.stackexchange.com/questions/2063091/prove-that-sum-limits-cyc4a65a5b-geq-fracabc627?noredirect=1&lq=1 — River Li, Jul 22 '20 at 01:59
My tool check this, $k_{\max} = \frac{31}{81}$ for the inequality $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge k (a+b+c)^6.$ https://i.imgur.com/TO4yJbI.png — NKellira, Aug 12 '20 at 23:09
@tthnew Very nice. So $\frac{40(a^6 + b^6 + c^6) + 53(a^5b+b^5c+c^5a)}{(a+b+c)^6}$ achieves its minimum at $a=b=c=1$. — River Li, Aug 13 '20 at 01:06
@RiverLi I can't not get SOS. It's very hard for me. But I can check inequality. — NKellira, Aug 13 '20 at 01:16
@tthnew I am not correcting your spelling. I cannot get SOS currently. — River Li, Aug 13 '20 at 02:06
@RiverLi Ok, my SOS is not true for all reals, it's only true when $a=b=c.$ https://i.imgur.com/eYQIc3p.png — NKellira, Aug 13 '20 at 02:13
@tthnew Nice. By the way, for $a, b, c\ge 0$, it is easy to prove the inequality by BW. So, if $a \ge b \ge c$, we get SOS, and for $b\ge a\ge c$, we get SOS. But your SOS is better. — River Li, Aug 13 '20 at 02:22
For $a \ge b \ge c$, let $b=c+s, a = c+s+t$ for $s, t \ge 0$ and we have $f(a,b,c)$ is a polynomial in $s, t$ with non-negative coefficients. Since $s = b - c, t = a-b$, we have $f(a,b,c) = (35720/9)c(b-c)^3*(a-b)^2 + \cdots$. It is not a standard SOS. Your SOS is standard. — River Li, Aug 13 '20 at 02:29
@nguyenhuyenag Could you give me P1, P3 SOS? I can't not do it by SOS. — NKellira, Feb 25 '21 at 06:04
@RiverLi I can give SOS for $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge \dfrac{31}{81}(a+b+c)^6$ but the coefficients is very very very and very big! — NKellira, Feb 25 '21 at 08:18

score 1 · Answer 1 · answered Feb 25 '21 at 08:14

My P2 SOS is ugly, the coefficients is too large, I will try to improve my tool:

We have:

$$27\left[4\left(a^6+b^6+c^6\right)+5\left(a^5b+b^5c+c^5a\right)\right]-(a+b+c)^6$$

$$={\frac {1}{418495910519665315875436}}{c}^{2} \left( 1877899493539b +2231128254114c \right) ^{2} \left( b-c \right) ^{2}$$

$$+{\frac {1}{1227564566709386794764}}\left( b-c \right) ^{2} \left( 111426599762{b}^{2}+57572325931bc-66554368268{c}^{2} \right) ^{2 }$$

$$+{\frac {1}{4982400602455205424}}\left( 11016800022a{c}^{2}- 8322034448{b}^{3}-16516031087{b}^{2}c+953764263b{c}^{2}+ 12867501250{c}^{3} \right) ^{2}$$

$$+{\frac {1}{110170254068928}}\left( 41114072abc-13516246\,a{c}^{2} -63487932{b}^{3}-24871895{b}^{2}c+68571471b{c}^{2}-7809470{c}^ {3} \right) ^{2}$$

$$+{\frac {1}{92668107698496}} \left( 58951728a{b}^{2}+20957464abc- 61903346a{c}^{2}+23285292{b}^{3}+20567219{b}^{2}c-22826379b{c} ^{2}-39031978{c}^{3} \right) ^{2}$$

$$+{\frac {1}{145642643664}}\left( 3143864{a}^{2}c-684660a{b}^{2}+ 1346196abc-252458\,a{c}^{2}-539136{b}^{3}-876131{b}^{2}c+54651b{c}^{2}-2192326{c}^{3} \right) ^{2}$$

$$+{\frac {1}{9913764}}\left( 23163{a}^{2}b+6338{a}^{2}c+11364a{ b}^{2}+3372abc-5897a{c}^{2}-19512{b}^{3}-9398{b}^{2}c-2028b{ c}^{2}-7402{c}^{3} \right) ^{2}$$

$$+{\frac {1}{428}} \left( 214{a}^{3}+129{a}^{2}b-6{a}^{2}c-108a{b}^{2}-56abc-87a{c}^{2}-8{b}^{3}-30{b}^{2}c-44b{c}^{2}-4 {c}^{3} \right) ^{2}+{\frac {5083151117301}{1877899493539}}{c}^{4} \left( b-c \right) ^{2}\ge 0,$$

which give us qed./

See also here.

@RiverLi SOStools of Matlab can't not give us this exact form, right. For P1 is very very ...large. — NKellira, Feb 25 '21 at 11:27
Yes, for small degree and small numbers of variables, the integers in rational SOS can be huge (> 100digits). — River Li, Feb 25 '21 at 11:29

score 1 · Answer 2 · answered Feb 25 '21 at 08:29

Since the exact SOS for the stronger problem of P1 is very very big, can not express in $\LaTeX;$

So here I only give the approximate SOS for $$40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge \dfrac{31}{81}(a+b+c)^6$$

Have

$$40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) - \dfrac{31}{81}(a+b+c)^6$$

$$\approx 39.62\, \left( {a}^{3}+ 0.6399\,{a}^{2}b- 0.02898\,{a}^{2}c- 0.5041\, a{b}^{2}- 0.3277\,abc- 0.4398\,a{c}^{2}+ 0.04222\,{b}^{3}- 0.08519\,{b }^{2}c- 0.2490\,b{c}^{2}- 0.04745\,{c}^{3} \right) ^{2}$$ $$+ 25.53\, \left( {a}^{2}c- 0.2476a{b}^{2}+ 0.4192abc- 0.04827a{c}^{2}- 0.1957\,{b}^{3}- 0.3349{b}^{2}c+ 0.03937b{c}^{2}- 0.6321{c}^{3} \right) ^{2}+ 17.98\left( {a}^{2}b+ 0.4437{a}^{2}c+ 0.4049a{b} ^{2}+ 0.1510\,abc- 0.1529\,a{c}^{2}- 0.8971\,{b}^{3}- 0.3941\,{b}^{2}c - 0.08513\,b{c}^{2}- 0.4704\,{c}^{3} \right) ^{2}+ 9.789\, \left( a{b} ^{2}+ 0.5075\,abc- 1.492\,a{c}^{2}+ 0.5095\,{b}^{3}+ 0.7374\,{b}^{2}c- 0.2611\,b{c}^{2}- 1.001\,{c}^{3} \right) ^{2}+ 6.027\, \left( abc+ 0.03308\,a{c}^{2}- 1.885\,{b}^{3}- 1.121\,{b}^{2}c+ 1.783\,b{c}^{2}+ 0.1908\,{c}^{3} \right) ^{2}+ 3.993\, \left( a{c}^{2}- 0.1356\,{b}^{3 }- 2.018\,{b}^{2}c- 0.7975\,b{c}^{2}+ 1.951\,{c}^{3} \right) ^{2}+ 0.06508\, \left( {b}^{3}- 0.6985\,{b}^{2}c- 1.349\,b{c}^{2}+ 1.047\,{ c}^{3} \right) ^{2}+ 0.03708\, \left( {b}^{2}c- 0.03128\,b{c}^{2}- 0.9687\,{c}^{3} \right) ^{2}+ 0.005711\, \left( b{c}^{2}- 1.0\,{c}^{3 } \right) ^{2}\ge 0,$$

which is obvious.

If you want to see the exact SOS, please visit Github.

score 1 · Answer 3 · answered Feb 05 '25 at 01:14

let $$f(a,b,c)=3240\sum a^6+4293\sum a^5b-31(a+b+c)^6$$

then we have $$\begin{aligned} f(a,b,c) & = \frac{5}{495800916786}\sum \left(- 9303539 a^{3} - 10857977 a^{2} b - 5472670 a^{2} c + 5316308 a b^{2} + 683176 a c^{2} + 9303539 b^{3} + 10174801 b^{2} c + 156362 b c^{2}\right)^{2} \\ & + \frac{11}{4157064178801396619778468002192}\left(\sum a b \left(- 13351379187648017 a + 6680328669020432 b + 6671050518627585 c\right)\right)^{2} \\ & + \frac{1}{1200953862413136}\left(\sum a \left(1325617592 a^{2} + 72598623 a b - 1001317224 b^{2} - 396898991 b c\right)\right)^{2} \\ & + \frac{10474449841008768165}{82524874758852393077}\left(\sum a b \left(b - c\right)\right)^{2} \end{aligned}$$

Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$

3 Answers3

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