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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

math110
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    The corresponding inequality for four variables is solved here: http://math.stackexchange.com/questions/714101/how-prove-this-inequality-fraca-sqrtab-sqrt3abc-sqrt4abcd4-le. – Martin R Apr 08 '16 at 07:43
  • You should post your solution as an answer, not add it to the question. Or what is your question now? – Martin R Apr 08 '16 at 07:47
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    See also: Kiran Kedlaya: Proof of a Mixed Arithmetic-Mean, Geometric-Mean Inequality (AMM, 1994) http://www.jstor.org/stable/2975630 http://www.dma.ulpgc.es/profesores/personal/aph/ficheros/resolver/ficheros/crp/mixed_1994.pdf – Martin Sleziak Apr 10 '16 at 11:17

1 Answers1

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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, {}]
NonalcoholicBeer
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