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.
- 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.
- 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.
Best Answer
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.