# For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt{abc}}$ hold?

By generalizing this (1) and this (2) questions and performing some research

$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt{abc}},\hbox{ for }a,b,c>0$$

for all $$0\le k.
The main goal was to prove the original inequality from (2), however, letting $$a=x^3,\,b=y^3,\,c=z^3$$ and clearing the denominator, the inequality becomes $$3 k x^3 y^3 z^3 + 3 \sum\limits_{sym}x^6 y^3 z^0 - \left(3+\frac k2\right)\sum\limits_{sym} x^5 y^2 z^2\ge 0\tag{1}$$ and I'm failing to apply Muirhead's inequality.
The method from this answer works only for $$k\le 3$$, and even with calculus I don't think that solving system of $$3$$ equations like $$\frac{\partial}{\partial x}$$LHS(1)$$=0$$: $$5 k x^3 y^2 z^2 - 9 k x y^3 z^3 + 2 k y^5 z^2 + 2 k y^2 z^5 - 18 x^4 y^3 - 18 x^4 z^3 + 30 x^3 y^2 z^2 - 9 x y^6 - 9 x z^6 + 12 y^5 z^2 + 12 y^2 z^5=0$$ may lead to something neat.)
Any help is appreciated. Thanks.

The question: what is $$k_0$$.

Let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, we need to find a maximal $$k$$ for which the following inequality is true for any positives $$a$$, $$b$$ and $$c$$. $$\frac{9uv^2}{w^3}+k-3\geq \left(2+\frac{k}{3}\right)\frac{3u}{w},$$ which says that it's enough to show it for a minimal value of $$v^2$$.

Now, $$a$$, $$b$$ and $$c$$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$3v^2x=-x^3+3ux^2+w^3.$$

Id est, the line $$y=3v^2x$$ and the graph of $$f(x)=-x^3+3ux^2+w^3$$ have three common points

(maybe less of three common points if this line is a tangent line to the graph).

We can draw a graph of $$f$$: $$f'(x)=-3x(x-2u),$$ which gives that $$(0,w^3)$$ is a minimum point and $$(2u,f(2u))$$ is a maximum point.

Now, we see that $$v^2$$ will get a minimal value, when $$y=3v^2x$$ would be a tangent to the graph of $$f$$,

which happens for equality case of two variables.

Since our inequality is homogeneous and symmetric we can assume $$b=c=1$$ and $$a=x^3$$, which gives $$\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\geq k,$$ which says $$k_0=\min_{x>0}\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\approx11.10864$$ Since $$\left(\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\right)'=\frac{6(x^2+x+1)(x^3+3x^2-3x-4)}{x^3(x+2)^2},$$ we see that this minimum occurs, when $$x$$ is a positive root of the equation: $$x^3+3x^2-3x-4=0,$$ which gives $$x_{min}=2\sqrt2\cos\left(\frac{1}{3}\arccos\left(-\frac{1}{4\sqrt2}\right)\right)-1.$$

• Thank you so much for describing in so much detail why it's the case of equality of two variables. I guess you guess I'm not familiar with the uvw method (I would had solved it otherwise maybe), that's why you wrote the details. Btw, do you mean "when $y=3v^2\color{blue}{x}$ would be a tangent"? Thanks. Jul 6, 2020 at 10:44
• @Alexey Burdin You are welcome! Yes, it was typo. I fixed. Jul 6, 2020 at 11:11

Suppose $$abc=1,$$ and let $$b=c=t,\,a=\frac{1}{t^2},$$ the inequality become $$k \leqslant \frac{6(t+1)(t^3+t^2+1)}{t(2t+1)} = F(t).$$ Easy to find $$k \leqslant k_0 = F(t_0) = \frac{9\sqrt{665}}{8}\sin{\left(\frac{\pi}{6}+\frac{1}{3}\arccos{\frac{13117\sqrt{665}}{442225}}\right)}-\frac{141}{16} = 11.1086,$$ for $$\displaystyle t_0 = \frac{\sqrt 5}{2} \cos \left(\frac{\arctan(2 \sqrt{31})}{3}\right)-\frac 14 = 0 .7345.$$

Finally, we will show that the inequality below is true for all $$k \leqslant k_0$$ $$f(a,b,c) = (a+b+c)(ab+bc+ca) + k - 3 - \left(\frac{k}{3}+2\right)(a+b+c) \geqslant 0.$$ Indeed, asumme $$a = \max \{a,b,c\}$$ and $$t = \sqrt{bc},$$ then $$a \geqslant 1,$$ we have $$f(a,b,c) - f(a,t,t) = (\sqrt b - \sqrt c)^2 \left[3(a^2+ab+bc+ca)+ 6 at - k - 6\right].$$ According to the AM-GM inequality, we have $$a^2+ab+bc+ca \geqslant 4a t,$$ so $$3(a^2+ab+bc+ca)+ 6 at \geqslant 18at \geqslant 18 > k_0+6 \geqslant k + 6.$$ Thefore $$f(a,b,c) \geqslant f(a,t,t),$$ and $$f(a,t,t) = f\left(\frac{1}{t^2},t,t\right) = \frac{(t-1)^2}{3t^3} \left[6(t+1)(t^3+t^2+1)-t(2t+1)k\right] \geqslant 0.$$ The proof is completed.

Note. This is Ji Chen inequality.