Find Min $P=\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}$ when $ab+bc+ca=3$

Question

$\forall a,b,c\ge 0: ab+bc+ca=3,$ find minimal value $$P=\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}$$ By $a=b=c=1$ we have $P\ge \frac{3}{\sqrt{2}}$ I try use AM-GM $$P^3\ge 27 \frac{1}{\sqrt{a+bc}}.\frac{1}{\sqrt{b+ca}}.\frac{1}{\sqrt{c+ab}}$$ We need prove $$2\sqrt{2}\ge \sqrt{(a+bc)(b+ca)(c+ab)}\iff (a+bc)(b+ca)(c+ab)\le 8$$Also by AM-GM $$(a+bc)(b+ca)(c+ab)\le \frac{(a+b+c+3)^3}{27}\le 8 \iff a+b+c\le 3$$ The last inequality is wrong.

How can I fix the approach? Thanks Updated edit:

How to find maximal value of $\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}$ ?

If we can find it, the miminum seems easy to get.

Updated: My aboved idea leads nothing since it is not true when $a=b\rightarrow 0.$

I also tried using Holder $$P^2.\sum(a+bc)(b+c)^3\ge 8(a+b+c)^3,$$ but the remain is something wrong. Whether Dragonboy's answer is eligible proof?

Answer 1

Proof.

By $a=b\rightarrow 0; c\rightarrow +\infty,$ we see that $P\rightarrow \dfrac{2\sqrt{6}}{3}.$

Thus, we'll prove that $$\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge \frac{2\sqrt{6}}{3}.$$ Remark. Honestly, the following proof is not nice but I think it is smooth enough by AM-GM and $uvw.$

Indeed, by using AM-GM it's enough to prove stronger inequality$$\frac{1}{2a+3+2bc}+\frac{1}{2b+3+2ca}+\frac{1}{2c+3+2ab}\ge \frac{1}{3}.\tag{*}$$ Let $a+b+c=3u; ab+bc+ca=3v^2; abc=w^3.$ The $(*)$ turns out $$f(w^3)=-4w^6+w^3(-36u^2+24u+20)+27u+18\ge 0. \tag{**}$$

Notice that: $f''(w^3)=-4<0,$ hence $f(w^3)$ is concave function on $w^3$ and the concave function gets a minimal value for an extremal value of $w^3$ , which by $uvw$ happens for the two following cases.

• $w^3=0:$ The $(**)$ is obviously true.
• Two variables are equal. Let $0<b=c=x\le \sqrt{3}; c=\dfrac{3-x^2}{2x}$, the $(**)$ becomes \begin{align*} &-4\left(\frac{3x-x^3}{2}\right)^2+\frac{3x-x^3}{2}\left[-4\left(2x+\frac{3-x^2}{2x}\right)^2+8\left(2x+\frac{3-x^2}{2x}\right)+20\right]+9\left(2x+\frac{3-x^2}{2x}+2\right)\\&=(-x^2 + 2 x + 6) \left(x^4 - \frac{5}{2}x^3 + x^2 + \frac{3}{2}x + 6\right)\ge 0, \forall 0<x\le \sqrt{3}. \end{align*}

Hence, $(**)$ is true and $(*)$ is proven. We end proof here.

About $uvw,$ see here. We can see also similar problem.

The $(**)$ can be solved by function and the following lower of $w^3$ $$w^3 \le \dfrac{3v^4(3u^2-v^2)}{6u(2u^2-v^2)}.$$

Answer 2

We will prove the inequality is true when $a=b\rightarrow 0$ or $$\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge \frac{2\sqrt{6}}{3}$$ Let $0\le a=b=t\le 1$ we get $c=\dfrac{1-t^2}{2t}$

Set limitation $$\lim_{x\rightarrow 0^{+}}\frac{1}{\sqrt{t+t\dfrac{1-t^2}{2t}}}+\frac{1}{\sqrt{t+t\dfrac{1-t^2}{2t}}}+\frac{1}{\sqrt{\dfrac{1-t^2}{2t}+t^2}}= \frac{2\sqrt{6}}{3} $$ We have Q.E.D

Updated editing

By squaring both side, it suffices to prove $$\frac{1}{a+bc}+\frac{1}{b+ac}+\frac{1}{c+ab}+2\sum_{cyc}\frac{1}{\sqrt{(a+bc)(b+ca)}}\ge \frac{8}{3}.$$ Does it exist lower bound of $$\frac{1}{a+bc}+\frac{1}{b+ac}+\frac{1}{c+ab}$$ and $$\sum_{cyc}\frac{1}{\sqrt{(a+bc)(b+ca)}}?$$

Answer 3

I drawed graph with variable $a,b$ letting $c=\frac{3-ab}{b+a}$ using software.

I also drawed graph when $c=\frac{3-ab}{b+a}, a=b=t$.

These graphs convince me that $P$ is indeed minimal around $a=b=0$ (or $b=c=0$, $c=a=0$), but this is not a formal proof.

(In my opinion, proving the inequality is hard because it is essentially $16$-degree polynomial in rational form.)

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[Edit] Just for my entertainment, I directly showed $f\geq 0$ with non-negative terms as following (with computer), where $f$ is described by @TATAbox

$f = (2uw^3-1)^2 + w^3(w^3-5u+4)^2/u+(w^3-6)^2/u+(w^3+9)g/u+6w^3+17$, where $g = -w^6+(6u-4u^3)w^3+3u^2-4 = ((a-b)(b-c)(c-a))^2 / 27$,