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

Answer 1

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.

Answer 2

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.