<s>[INST] prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods  links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
 [/INST] Problem: Let $a, b, c > 0$ with $a^2+b^2+c^2+abc=4$. Prove that $a^3+b^3+c^3+(abc)^3\ge 4$.
Solution: It suffices to prove that, for $a, b, c > 0$,
$$a^3+b^3+c^3+(abc)^3 - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{1}$$
It is verified by Mathematica.
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when
$a=b$. The proof is ugly. Omitted. If I find nice proofs, I will update the post.
I hope to see simple SOS (Sum of Squares) solutions of (1).
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
% https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Remark: The stronger version is also true: for $a, b, c > 0$,
$$a^3+b^3+c^3+2(abc)^2 - abc - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{2}$$
(Note: $(abc)^3 + abc \ge 2(abc)^2$.)
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