0

When I'm trying this problem, I find it could be harder if choose a larger $k$ (in the original problem $k=13/5=2.6$):

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^3}{ka^2+b^2}+\frac{b^3}{kb^2+c^2}+\frac{c^3}{kc^2+a^2}\geq\frac{a+b+c}{k+1}$$

e.g. the inequality still holds when $k=164/63=2.6032...$ (I proved it by Successive Difference Substitution, a computer-automated buffalo-way. I think it can also be proved by CAD or other methods mentioned in the original thread) but doesn't hold when $k=315/121=2.6033...$ (the figure shows 3 very small negative areas).

k=315/121

I guess the critical point is just a slight above 2.60325... (still holds at $2.603250$), but I don't know if it is a rational number or irrational (or transcendental).

auntyellow
  • 579
  • It's most definitely a root of some extremely hairy polynomial (with radicals allowed probably). Much simpler question like $x^3+y^3+z^3+k(xy^2+yz^2+zx^2)\geq(k+1)(x^2y+y^2z+z^2x)$ turns out to have the optimal (in this case maximal) $k$ to be a root of a cubic, for example. – dezdichado Jan 24 '24 at 22:28
  • By Mathematica, the maximal $k\approx 2.603260274$ is a real root of $64,{x}^{15}-1504,{x}^{14}+13076,{x}^{13}-38833,{x}^{12}-49126,{x} ^{11}+146301,{x}^{10}+316414,{x}^{9}-466451,{x}^{8}-61696,{x}^{7}+ 1051234,{x}^{6}+162556,{x}^{5}+290137,{x}^{4}+768982,{x}^{3}+ 157577,{x}^{2}+262722,x+258147 =0$. – River Li Jan 25 '24 at 02:34
  • @dezdichado in your question the critical point is $k^4+2k^3-5k^2-6k-23=0$ then $k=(\sqrt{13+16\sqrt{2}}-1)/2=2.4844...$. – auntyellow Jan 25 '24 at 09:32
  • @RiverLi can you provide the detail how you (or Mathematica) get the 15-degree equation? – auntyellow Jan 25 '24 at 09:38
  • 1
    @auntyellow Mathematica code: f[a_, b_] := a^5b^2k + a^4b^3k - a^4b^2k^2 - a^3b^4k + a^3b^2k^2 + a^2b^5 - a^2b^4k + a^2b^3k^2 - ab^4k^2 - a^4bk + a^2b^2k^2 + b^5k + a^5 - a^4k - a^3b^2 - a^2b^3 - a^2bk^2 + b^4k + a^3k - a^2b^2 - ab^2k - b^3k + a^2k + b^2 pos1 := a >= 0 && b >= 0 Resolve[ForAll[{a, b}, pos1, f[a, b] >= 0], Reals] (Let $c = 1$ in the original inequality and clear the denominators. For the code, see this) – River Li Jan 25 '24 at 12:38

0 Answers0