If $a,b,c\ge 0: ab+bc+ca=3$ then prove that: $$\sqrt{2a+3bc}+\sqrt{2b+3ca}+\sqrt{2c+3ab}\ge 3\sqrt{5}.$$
I've tried to use AM-GM and Holder without success.
Indeed, by AM-GM for three numbers $$\sqrt{2a+3bc}+\sqrt{2b+3ca}+\sqrt{2c+3ab}\ge 3\sqrt[6]{(2a+3bc)(2b+3ca)(2c+3ab)}$$ and we need to prove $$(2a+3bc)(2b+3ca)(2c+3ab)\ge 125$$ I check it is wrong when $a=b=\sqrt{3}; c=0.$
Also by Holder, $$\left(\sum_{cyc}\sqrt{2a+3bc}\right)^2\cdot \sum_{cyc}\frac{(b+c+xa)^3}{2a+3bc}\ge (x+2)^3(a+b+c)^3$$ I was stuck to find $x$ satisfying $$(x+2)^3(a+b+c)^3\ge 45\cdot \sum_{cyc}\frac{(b+c+xa)^3}{2a+3bc}$$ since equality occurs when $a=b=c=1.$
I think there is a better Holder using but I've not found it so far.
Update $1:$ I tried AM-GM $$\sqrt{2a+3bc}=\frac{2a+3bc}{\sqrt{2a+3bc}}\ge \frac{2(2a+3bc)}{\dfrac{2a+3bc}{a+tbc}+a+tbc}$$but it's not smooth.
Proof.
The desired inequality is written as $$\sqrt{\frac{2a + 3bc}{5}} + \sqrt{\frac{2b + 3ca}{5}} + \sqrt{\frac{2c + 3ab}{5}} \ge 3. \tag{1}$$
By AM-GM, we have $$\sqrt{\frac{2a + 3bc}{5}} = \frac{12(2a + 3bc)}{2\sqrt{180(2a + 3bc)}} \ge \frac{12(2a + 3bc)}{\frac{180(2a + 3bc)}{25 + 5bc} + (25 + 5bc)}. \tag{2}$$
From (1) and (2), it suffices to prove that $$\sum_{\mathrm{cyc}} \frac{12(2a + 3bc)}{\frac{180(2a + 3bc)}{25 + 5bc} + (25 + 5bc)} \ge 3. \tag{3}$$
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$. We have $$p \ge 3, \quad 0 \le r \le 1, \quad r \ge \frac{12p - p^3}{9}. \tag{4}$$ (Note: $r \ge \frac{12p - p^3}{9}$ follows from the degree three Schur inequality $r \ge \frac{4pq - p^3}{9}$.)
(3) is equivalently written as $$775r^4 + m_3 r^3 + m_2 r^2 + m_1 r + m_0 \ge 0 \tag{5}$$ where \begin{align*} m_3 &= 200p^2 + 16510p + 3360, \\ m_2 &= 7840p^3 + 6115p^2 - 4720p + 672952, \\ m_1 &= 788768p^2 - 1388944p - 3990360, \\ m_0 &= 16000p + 29308. \end{align*}
It suffices to prove that (5) is true for all $p, r$ satisfying (4).
Since $m_3, m_2, m_0 \ge 0$, we only need to prove the case that $m_1 < 0$ which implies $p < 2\sqrt{3}$. Thus, $r \ge \frac{12p - p^3}{9} \ge 0$. Let $r_1 := \frac{12p - p^3}{9}$. We have \begin{align*} &775r^4 + m_3 r^3 + m_2 r^2 + m_1 r + m_0\\ \ge{}& 775r_1^4 + m_3 r_1^3 + m_2 r_1^2 + m_1 + m_0 \\ ={}& \frac{1}{6561}(p - 3)g(p)\\ \ge{}& 0 \end{align*} where \begin{align*} g(p) &:= 775\,{p}^{11}+525\,{p}^{10}-184215\,{p}^{9}+116955\,{p}^{8}+6865020\,{ p}^{7}\\ &\qquad +5282820\,{p}^{6}-11077668\,{p}^{5}+57435156\,{p}^{4}-791753940 \,{p}^{3}\\ &\qquad -2378061180\,{p}^{2}+5890235436\,p+8662820724. \end{align*}
We are done.