Let $a,b,c\ge 0: a+b+c+abc=4$. Prove that$$\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge \frac{2\sqrt{2}+1}{2}.$$
I've try to use AM-GM which is very complicated.
Indeed, we can use the below estimate $$\frac{1}{\sqrt{a+bc}}\ge \frac{\sqrt{2}a+bc}{\frac{1}{2}a^2+\left(\frac{\sqrt{2}}{2}bc+1\right)a+bc+\frac{1}{4}(bc)^2}=4\frac{\sqrt{2}a+bc}{2a^2+\left(2\sqrt{2}bc+4\right)a+4bc+(bc)^2},$$and we need to prove$$\sum_{\mathrm{cyc}}{\frac{\sqrt{2}a+bc}{2a^2+\left(2\sqrt{2}bc+4\right)a+4bc+(bc)^2}}\ge \frac{2\sqrt{2}+1}{8}.$$
I check it is true when $a=b$ and $c=\dfrac{4-2a}{a^2+1}.$
Also, I tried Holder inequality as below}$$\left(\sum_{cyc}\frac{1}{\sqrt{a+bc}}\right)^2.\sum_{cyc}(a+bc)[(2\sqrt{2}-1)a+b+c]^3\ge(2\sqrt{2}+1)^3(a+b+c)^3 ,$$which saves the equality cases but it's wrong when $0\le a=b=x<0.503(!).$
Is there a simple approach to apply ? I'm waiting for a handle proof which students can full it in during contest time.
Thanks for interest!
Sketch of a proof.
Remarks: Naturally, I deal with it using AM-GM and C-S. (Fortunately, it is true.)
WLOG, assume that $a \ge b \ge c$.
By AM-GM, it suffices to prove that $$\frac{2\sqrt 2}{a + bc + 2} + \frac{2\sqrt 2}{b + ca + 2} + \frac{4}{c + ab + 4} \ge \sqrt 2 + \frac12.$$
Using Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{2\sqrt 2}{a + bc + 2} + \frac{2\sqrt 2}{b + ca + 2} \ge 2\sqrt{2} \cdot \frac{4}{a + bc + 2 + b + ca + 2}.$$
It suffices to prove that $$2\sqrt{2} \cdot \frac{4}{a + bc + 2 + b + ca + 2} + \frac{4}{c + ab + 4} \ge \sqrt 2 + \frac12.$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\left(2\sqrt{2} \cdot \frac{4}{a + bc + 2 + b + ca + 2} + \frac{4}{c + ab + 4}\right)\\[6pt] &\quad \times \left(\sqrt{2}\, (a + bc + 2 + b + ca + 2) + \frac12(c + ab + 4)\right)\\[6pt] \ge{}& (4 + \sqrt{2})^2. \end{align*}
It suffices to prove that $$\frac{(4 + \sqrt{2})^2}{\sqrt{2}\, (a + bc + 2 + b + ca + 2) + \frac12(c + ab + 4)} \ge \sqrt{2} + \frac12$$ or $$2 + 4\sqrt{2} - \sqrt{2}\, c(a + b) - \sqrt{2}(a + b) - \frac12ab - \frac12 c \ge 0. \tag{1}$$
Using $c = \frac{4 - a - b}{1 + ab}$, (1) is equivalently written as \begin{align*} f(b) &:= - ( 2\sqrt {2}\, a + {a}^{2} - 2\sqrt {2}) {b}^{2}+ ( -2 \sqrt {2}\,{a}^{2} + 12\sqrt {2}\, a - 10\sqrt {2} + 3a+1) b\\ &\qquad +2\, \sqrt {2}\,{a}^{2}-10\,\sqrt {2}\, a+8\,\sqrt {2}+a\\ &\ge 0. \end{align*}
We have $$1 \le a \le 4, \quad 1 - \frac{a}{4} \le b \le 4 - a.$$ (Note: From $b \ge c = \frac{4 - a - b}{1 + ab}$, we have $b \ge \frac{\sqrt{1 + a(4 - a)} - 1}{a} \ge \frac{1 + a(4 - a)/4 - 1}{a} = 1 - \frac{a}{4}$.)
Since $a\ge 1$, clearly $f(b)$ is concave. We have $f(1 - a/4)\ge 0$ and $f(4 - a)\ge 0$ for all $1 \le a \le 4$. Thus, $f(b) \ge 0$ for all $1 - a/4 \le b \le 4 - a$, given any $1\le a\le 4$.
We are done.