Inequality $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\leq \frac{\sqrt{3}}{k+1}$

Question

Hi it's related to this If $a+b+c = 3abc$ and $\frac17 \leq k \leq 7$ prove $ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $

I propose this :

Let $a,b,c>0$ and $a+b+c=abc$ and $a\geq b \geq c $ then we have : $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\leq \frac{\sqrt{3}}{k+1}$$ Where $k$ is a real numbers such that $8\leq k\leq \alpha $ with : $$\alpha^6 - 12 \alpha^5 + 30 \alpha^4 - 42 \alpha^3 + 30 \alpha^2 - 12 \alpha + 1=0 $$

I try a lot of things Karamata's inequality by example . More interesting I think we can use strong convexity . In the case of $k=8$ we have :

$$\frac{1}{8a+b}\leq \frac{1}{9}\Big(\frac{8}{9a}+\frac{1}{9b}-\frac{8}{81}\frac{(a-b)^2}{a^3}\Big)$$ And $$\frac{1}{8b+c}\leq \frac{1}{9}\Big(\frac{8}{9b}+\frac{1}{9c}-\frac{8}{81}\frac{(b-c)^2}{b^3}\Big)$$

Remains to show with the initial conditions: $$\frac{1}{9}\Big(\frac{8}{9b}+\frac{1}{9c}-\frac{8}{81}\frac{(b-c)^2}{b^3}\Big)+\frac{1}{9}\Big(\frac{8}{9a}+\frac{1}{9b}-\frac{8}{81}\frac{(a-b)^2}{a^3}\Big)+\frac{1}{8c+a}\leq\frac{\sqrt{3}}{9}$$

But I think it's not true and we lost some cyclicity . So I'm really stuck .

Maybe Buffalo way's can kills it but I don't know how to use it .

Thanks a lot for sharing your time and knowledge .

Answer 1

The Buffalo Way works.

After homogenization, it suffices to prove that, for $a\ge b\ge c > 0$ and $8 \le k\le \alpha$, $$\frac{3}{(k+1)^2}\ge \frac{abc}{a+b+c}\left(\frac{1}{ka+b} + \frac{1}{kb+c} + \frac{1}{kc+a}\right)^2.$$ WLOG, assume that $c = 1$. Let $a = 1 + s + t, b = 1 + s$ for $s, t\ge 0$. It suffices to prove that, for $s, t \ge 0$ and $8 \le k\le \alpha$, $$\frac{3}{(k+1)^2}\ge \frac{(1+s+t)(1+s)}{3+2s+t}\left(\frac{1}{k(1+s+t)+1+s} + \frac{1}{k(1+s)+1} + \frac{1}{k+1+s+t}\right)^2.$$ After clearing the denominators, it suffices to prove that, for $s, t \ge 0$ and $8 \le k\le \alpha$, $$q_5t^5 + q_4t^4 + q_3t^3 + q_2t^2 + q_1t + q_0 \ge 0$$ where $q_5, q_4, q_3, q_2, q_1, q_0$ are polynomials in $(s, k)$.

It is easy to prove that $q_5, q_4, q_3, q_2, q_1\ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$. Indeed, for $8 \le k\le \alpha$, each of them can be expressed as polynomials in $s$ with non-negative coefficients (polynomials in $k$).

Thus, it suffices to prove that $q_0 \ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$. We have \begin{align} q_0 &= 6 k^2 s^3+(-k^4+6 k^3+22 k^2+6 k-1) s^2\\ &\quad +(-2 k^4+14 k^3+32 k^2+14 k-2) s -(k^2-10k+1)(k+1)^2. \end{align} For each fixed $k$ with $8 \le k\le \alpha$, $q_0 = q_0(s)$ is a cubic function of $s$. Note that $q_0(0) = -(k^2-10k+1)(k+1)^2 > 0$, $q_0(-\infty) = -\infty$ and $q_0(\infty) = \infty$ for $8 \le k\le \alpha$. Also, for $8 \le k\le \alpha$, $q_0(s)$ has a discriminant \begin{align} \mathrm{discr}(q_0) &= 12 k^2 (k^6-12 k^5+30 k^4-42 k^3+30 k^2-12 k+1) (k+1)^2 (k-1)^4\\ &\le 0. \end{align} As a result, we have $q_0 \ge 0$ for $s\ge 0$ and $8 \le k\le \alpha$.

We are done.