<s>[INST] Prove with some AM-GM inequality? I have proved the following inequality:
Let $a,b,c>0$
$$\dfrac{(a+\sqrt{ab}+\sqrt[3]{abc})}{3}\le \sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$
My solution is:$$a\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}   =\dfrac{1}{3^3}(a+a+a)(a+\dfrac{a+b}{2}+b)(a+b+c)\ge\dfrac{1}{3^3}(a+a+a)(a+\sqrt{ab}+b)(a+b+c)\ge \dfrac{1}{3}(a+\sqrt{ab}+\sqrt[3]{abc})^3$$
But I met the following, harder inequality:
$$3(a+\sqrt{ab}+\sqrt[3]{abc})\le \left(7+\dfrac{4\sqrt{ab}}{a+b}\right)\sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$
How to prove this inequality? Thanks
 [/INST] This is a proof by computer.
Substitute by $a=x^3, b=y^3, c=z^3$, the original inequality is equivalent to
show a polynomial in $x,y,z$ of degree $30$ is non-negative. A Cylindrical Decomposition shows this is true.
The code is the following
ieq1 = 3 (a + Sqrt[a b] + (a b c)^(1/3)) <= (7 + (4 Sqrt[a b])/(
      a + b)) (a (a + b)/2 (a + b + c)/3)^(1/3);

assume = a >= 0 && b >= 0 && c >= 0;

f1[a_, b_, c_] = 
  ieq1[[2]]^3 - ieq1[[1]]^3 // Together // Numerator // 
   Simplify[#, assume] &;

assume1 = x >= 0 && y >= 0 && z >= 0;

f2[x_, y_, z_] = f1[x^6, y^6, z^6] // Simplify[#, assume1] &;

ieq2 = ForAll[{x, y, z}, assume1, f2[x, y, z] >= 0];

CylindricalDecomposition[ieq2, {}]

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