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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.

  1. 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.

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  • $\begingroup$ Have you tried the method of Lagrange Multipliers or is it not applicable? $\endgroup$
    – Tran Khanh
    Oct 25, 2023 at 4:45
  • $\begingroup$ It seems hard to apply $\endgroup$
    – TATA box
    Oct 25, 2023 at 7:08
  • $\begingroup$ Try homogenation with $ab+bc+ca = 1$ $\endgroup$
    – Krave37
    Oct 26, 2023 at 6:55

1 Answer 1

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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.

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  • $\begingroup$ Very nice, thank you! $\endgroup$
    – TATA box
    Oct 26, 2023 at 13:15
  • $\begingroup$ @TATAbox It works for your another problem. I just posted an answer for that. $\endgroup$
    – River Li
    Oct 26, 2023 at 13:22
  • $\begingroup$ I like it! It is also here: math.stackexchange.com/questions/4740147/… $\endgroup$
    – TATA box
    Oct 26, 2023 at 14:01
  • $\begingroup$ @TATAbox oh, yes. $\endgroup$
    – River Li
    Oct 26, 2023 at 14:04
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    $\begingroup$ @MichaelRozenberg Thanks for checking. uvw is more terse. $\endgroup$
    – River Li
    Oct 27, 2023 at 5:27

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