Let $a,b,c$ be the side lengths of a triangle. How to prove the following inequality? $$\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}+\frac{\sqrt{ac}\,ac}{b(a+b+c)(a+c-b)}+\frac{\sqrt{ab}\,ab}{c(a+b+c)(a+b-c)}\geq1$$
Via some numerical experiment and graphing techniques, I am sure it is true. Could we use AM-GM?
My goal is to find an analytic/conventional proof of this inequality. I have tried AM-GM, but it didn't work. Since it is a homogeneous function with degree 0, WLOG, let $a=1.$ Then I plotted the LHS of this inequality using the Maple command "plot3d" which clearly shows that the LHS ≥ 1.
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\begin{align} \f abc+\f bca+\f cab &\ge 1 \tag{1}\label{1} \end{align}
By AM-GM
\begin{align} &\f abc+\f bca +\f cab \\ &\ge 3\sqrt[3]{ \f abc\cdot\f bca\cdot\f cab } \tag{2}\label{2} \\ &= 3\sqrt[3]{ \frac{(abc)^2}{ (a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) } } \tag{3}\label{3} . \end{align}
Let $\rho$, $r$ and $R$ be the semiperimeter, inradius and circumradius of corresponding triangle.
Then we can rewrite \eqref{3} as
\begin{align} &3\sqrt[3]{ \frac{(4\rho r R)^2}{ 16(2\rho)^2\rho(\rho-a)(\rho-b)(\rho-c)}} \tag{4}\label{4} \\ &= 3\sqrt[3]{ \frac{(\rho r R)^2}{ (2\rho)^2\rho^2r^2 } } =3\sqrt[3]{\frac{R^2}{4\rho^2}} =\frac{3}{\sqrt[3]{4(\rho/R)^2}} \tag{5}\label{5} . \end{align}
All of the above statements equally hold for a similar normalized triangle, scaled by $\tfrac1R$, so let $u=\rho/R$ and $v=r/R$. This parameterization is useful, since all possible valid triangular shapes are uniquely defined by a pair $v,u$ for all $v\in[0,\tfrac12]$. Also, for any given $v$ all the valid values of $u$ are located between
\begin{align} u_{\min}&=\sqrt{27-(5-v)^2-2\sqrt{(1-2\,v)^3}} \tag{6}\label{6} ,\\ u_{\max}&=\sqrt{27-(5-v)^2+2\sqrt{(1-2\,v)^3}} \tag{7}\label{7} \end{align}
and of course, $u_{\max}\ge u_{\min}$ for all valid values of $v$.
So,
\begin{align} \frac{3}{\sqrt[3]{4(\rho/R)^2}} &= \frac{3}{\sqrt[3]{4\,u^2}} \ge \frac{3}{\sqrt[3]{4\,u_{\max}(v)^2}} \ge \frac{3}{\sqrt[3]{4\,\Big(\displaystyle\max_{v\in[0,\tfrac12]}u_{\max}(v)\Big)^2}} \tag{8}\label{8} . \end{align}
Note that $u_{\max}(v)$ is increasing on $v=[0,\tfrac12]$ and
\begin{align} \max_{v\in\Big[0,\tfrac12\Big]}u_{\max}(v) &= u_{\max}(\tfrac12) =\tfrac32\,\sqrt3 \tag{9}\label{9} , \end{align}
so we have
\begin{align} \frac{3}{\sqrt[3]{4\,\Big(\displaystyle\max_{v\in[0,\tfrac12]}u_{\max}(v)\Big)^2}} &= \frac{3}{\sqrt[3]{4\,\Big(\tfrac32\,\sqrt3\Big)^2}} =1 \tag{10}\label{10} , \end{align}
and the proof is complete.
It looks like that some comment on the Eqns \eqref{6}, \eqref{7} is due. As it was mentioned in the comments, one reference can be found in, for example, p.2, Eqs.(2),(3) in Mitrinovic, D.S., Pecaric, J. and Volenec, V., 1989. Recent advances in geometric inequalities (Vol. 28). Brill Archive. More references can be found there, one is as old as of 1890-1891. But basically, it follows from well-known expressions:
\begin{align} a_2&=a+b+c=2\rho \tag{11}\label{11} ,\\ a_1&=ab+bc+ca=\rho^2+r^2+4rR \tag{12}\label{12} ,\\ a_0&=abc=4\rho r R \tag{13}\label{13} \end{align} and hence, the roots of this cubic equation \begin{align} x^3-a_2 x^2+a_1 x-a_0&=0 \tag{14}\label{14} \end{align} are the three side length $a,b,c$ of the triangle with given $\rho,r$ and $R$.
Roots of \eqref{14} are all real, when its discriminant
\begin{align} \Delta(1,-a_2,a_1,-a_0)&= 18\,a_2\,a_1\,a_0-4\,a_2^3\,a_0+a_2^2\,a_1^2-4\,a_1^3-27\,a_0^2 \\ &= -4\,r^2\,(\rho^4 -(4\,R\,(5\,r+R)-2\,r^2)\,\rho^2+r\,(r+4\,R)^3) \ge 0 \tag{15}\label{15} , \end{align} and equations \eqref{6}, \eqref{7} follow.
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