Showing $\frac{1}{5}<(f(k,l,m,n))^3\leq\frac{64}{11}$, where $f(k,l,m,n)=\sum_{cyc}\frac{k^2}{\sqrt[3]{5k^6+6l^3m^3}}$ for positive $k$, $l$, $m$, $n$

Question

For $k,l,m,n \in \mathbb{R}^+$ denote by $f(k,l,m,n)$ the following function: $$f(k,l,m,n) = \frac{k^2}{\sqrt[3]{5k^6+6l^3m^3}}+\frac{l^2}{\sqrt[3]{5l^6+6m^3n^3}}+\frac{m^2}{\sqrt[3]{5m^6+6n^3k^3}}+\frac{n^2}{\sqrt[3]{5n^6+6k^3l^3}}$$ Prove that: $$\frac{1}{5}<(f(k,l,m,n))^3\leq\frac{64}{11}$$

What I Tried: First I got the equality case for the right side inequality, which holds at $k = l = m = n = c$ for any constant $c \in \mathbb{R}^+$.

Now, the next thing is I spotted that the inequalities are homogenous, so its possible to normalize the terms and directly assuming $klmn = 1$ seems like a good option. I haven't been able to properly use it though.

Next bit, all terms belonging to positive reals gives us a pretty good hint that AM-GM is lurking in some step. But for this I directly used AM-GM first on the $11$ terms of a denominator (its necessary to hold the equality case so I had to do $11$ terms) which gives us:

$$5k^6 + 6l^3m^3 \geq 11\sqrt[11]{k^{30}l^{18}m^{18}}$$ So, if we can perhaps show, $$\implies f(k,l,m,,n) \leq \sum_{cyc} \frac{k^2}{11\sqrt[11]{k^{30}l^{18}m^{18}}} \leq \frac{4}{\sqrt[3]{11}}$$ that this is true somehow, then we will be $1$ side done to the inequality. Also, the structure of the inequality makes it feel like holder's also a possibility, but I couldn't quite apply it here.

Could someone help me out?

Answer 1

Problem. Let $a, b, c, d > 0$. Prove that $$\sqrt[3]{1/5} < \sum_{\mathrm{cyc}(a,b,c,d)} \frac{a^2}{\sqrt[3]{5a^6 + 6b^3c^3}} \le \sqrt[3]{64/11}.$$

Sketch of a proof.

• For the left inequality

Using $6b^3c^3 = \frac34(2bc)^3 \le \frac34(b^2 + c^2)^3 \le 5 (b^2 + c^2 + d^2)^3$, we have \begin{align*} \sqrt[3]{5a^6 + 6b^3c^3} &\le \sqrt[3]{5a^6 + 5 (b^2 + c^2 + d^2)^3}\\ &= \sqrt[3]{5}\sqrt[3]{a^6 +(b^2 + c^2 + d^2)^3}\\ &\le \sqrt[3]{5}(a^2 + b^2 + c^2 + d^2). \end{align*}

We have $$\sum_{\mathrm{cyc}(a,b,c,d)} \frac{a^2}{\sqrt[3]{5a^6 + 6b^3c^3}} \ge \sum_{\mathrm{cyc}(a,b,c,d)} \frac{a^2}{\sqrt[3]{5}(a^2 + b^2 + c^2 + d^2)} = \sqrt[3]{1/5}.$$ Clearly, equality does not occur.

$\phantom{2}$

• For the right inequality

The right inequality is written as $$\sum_{\mathrm{cyc}(a,b,c,d)} \frac{1}{\sqrt[3]{5 + 6b^3c^3/a^6}} \le \sqrt[3]{64/11}.$$

Let $$x := b^3c^3/a^6, y := c^3d^3/b^6, z := d^3a^3/c^6, w := a^3b^3/d^6.$$ Then $xyzw = 1$.

It suffices to prove that, for all $x, y, z, w > 0$ with $xyzw = 1$, $$\frac{1}{\sqrt[3]{5 + 6x}} + \frac{1}{\sqrt[3]{5 + 6y}} + \frac{1}{\sqrt[3]{5 + 6z}} + \frac{1}{\sqrt[3]{5 + 6w}} \le \sqrt[3]{64/11}. \tag{1}$$

(1) is true which is verified by Mathematica. I have a proof below.

WLOG, assume that $x \ge y \ge z \ge w$. We split into three cases.

Case 1: $x < 10$

It is not difficult to prove that $$\frac{1}{\sqrt[3]{5+6u}} \le \frac{1}{\sqrt[3]{11}} - \frac{2\sqrt[3]{121}}{121}\ln u, \quad \forall u \in (0, 10).$$

Thus, we have $$\mathrm{LHS}_{(1)} \le \frac{4}{\sqrt[3]{11}} - \frac{2\sqrt[3]{121}}{121}\ln (xyzw) = \sqrt[3]{64/11}.$$

Case 2: $y \le 5/2$

Let $g(u) := \frac{1}{\sqrt[3]{5 + 6\mathrm{e}^u}}$. We have $g''(u) = \frac{2\mathrm{e}^u(2\mathrm{e}^u - 5)}{(5 + 6\mathrm{e}^u)^{7/3}}$. Thus, $g(u)$ is concave on $(-\infty, \ln\frac52]$. By Jensen's inequality, we have \begin{align*} \mathrm{LHS}_{(1)} &= \frac{1}{\sqrt[3]{5 + 6x}} + g(\ln y) + g(\ln z) + g(\ln w)\\ &\le \frac{1}{\sqrt[3]{5 + 6x}} + 3 \cdot g\left(\frac{\ln y + \ln z + \ln w}{3}\right)\\ &= \frac{1}{\sqrt[3]{5 + 6x}} + 3 \cdot \frac{1}{\sqrt[3]{5 + 6\sqrt[3]{yzw}}}\\ &= \frac{1}{\sqrt[3]{5 + 6x}} + 3 \cdot \frac{1}{\sqrt[3]{5 + 6\sqrt[3]{1/x}}}. \end{align*} The rest is smooth.

Case 3: $x \ge 10$ and $y > 5/2$

We have $$\mathrm{LHS}_{(1)} \le \frac{1}{\sqrt[3]{5 + 6\cdot 10}} + \frac{1}{\sqrt[3]{5 + 6\cdot 5/2}} + \frac{1}{\sqrt[3]{5}} + \frac{1}{\sqrt[3]{5 }} < \sqrt[3]{64/11}.$$