If $ab+bc+ca=1,$ prove that $\frac{\sqrt{5a+4}}{a+bc}+\frac{\sqrt{5b+4}}{b+ca}+\frac{\sqrt{5c+4}}{c+ab}\ge 8.$

Question

Problem. Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that $$\frac{\sqrt{5a+4}}{a+bc}+\frac{\sqrt{5b+4}}{b+ca}+\frac{\sqrt{5c+4}}{c+ab}\ge 8.$$

---

I've tried to use Holder inequality without success.

$$\left(\sum_{cyc}\frac{\sqrt{5a+4}}{a+bc}\right)^2.\sum_{cyc}(5a+4)^2(a+bc)^2\ge [5a+5b+5c+12]^3. \tag{1}$$

$$\left(\sum_{cyc}\frac{\sqrt{5a+4}}{a+bc}\right)^2.\sum_{cyc}(5a+4)^2(a+bc)^2(b+c)^3\ge \left(\sum_{cyc}(5a+4)(b+c)\right)^3.\tag{2} $$

Which are both leads to wrong inequalities in general.

Also, a big trouble here is equality case. It is $(a,b,c)=(0,1,1)$ but when I denote $a=b\rightarrow 0$, the RHS is approximate to $8.$

I'd like to ask two questions.

1. Is there a better Holder using ?

I think the appropriate one might be ugly but if you find it, please feel free to share it here.

2. Are there others idea which are smooth enough?

For example, Mixing variables, AM-GM or Cauchy-Schwarz...etc.

I aslo hope to see a good lower bound of $\frac{\sqrt{5a+4}}{a+bc},$ which eliminates the radical form (may be the rest is simpler by $uvw$)

All ideas and comments are welcomed. Thank you for your interest!

Remark. About $uvw$, see here.

Answer 1

Sketch of a proof.

By AM-GM, we have $$\frac{\sqrt{5a+4}}{a+bc} = \frac{2(5a+4)}{(a+bc)\cdot 2\sqrt{5a + 4}} \ge \frac{2(5a+4)}{(a+bc)\cdot \left(\frac{5a + 4}{a + 2} + a + 2\right)}.$$

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2(5a+4)}{(a+bc)\cdot \left(\frac{5a + 4}{a + 2} + a + 2\right)} \ge 8. \tag{1}$$

We use the pqr method. Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.

(1) is written as \begin{align*} f(r) &:= -8\,{r}^{4}+ \left( -8\,{p}^{2}-494\,p-4384 \right) {r}^{3}\\ &\qquad + \left( - 520\,{p}^{3}-3122\,{p}^{2}+2408\,p-8026 \right) {r}^{2}\\ &\qquad + \left( -512\, {p}^{4}-3680\,{p}^{3}+2412\,{p}^{2}+17786\,p+1508 \right) r\\ &\qquad +128\,{p}^{ 3}+992\,{p}^{2}-512\,p-3968\\ &\ge 0. \tag{2} \end{align*}

We have $f''(r) = -96\,{r}^{2}+6\, \left( -8\,{p}^{2}-494\,p-4384 \right) r-1040\,{p}^{3 }-6244\,{p}^{2}+4816\,p-16052 < 0$. Thus, $f(r)$ is concave.

By degree three Schur inequality, we have $r \ge \frac{4pq - p^3}{9} = \frac{p(2 - p)(2 + p)}{9}$. Thus, we have $$r \ge \max\left(0, \, \frac{p(2 - p)(2 + p)}{9}\right) =: r_1.$$

Also, from $0 \le (a-b)^2(b-c)^2(c-a)^2 = -4\,{p}^{3}r+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2}$, we have $$r \le - \frac{2}{27} p^3 + \frac{p}{3} + \frac{2}{27}(p^2 - 3)\sqrt{p^2 - 3} =: r_2.$$

We can prove that $f(r_1) \ge 0$ and $f(r_2) \ge 0$. Thus, we have $f(r) \ge 0$ on $[r_1, r_2]$. The desired result follows.