quykhtn-qa1 wrote:

$\triangle\, ABC\ \implies\ \boxed{\ \frac 1{l_a}+\frac 1{l_b}+\frac 1{l_c}\, \ge\, \frac {\sqrt{\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}}{abc\sqrt 2}\ }$

Proof. Firstly, let us note that: $\boxed{\sum\, \frac 1{l_a}}\stackrel{(1)}{=}\frac 1{2rs}\sum\, \frac {ah_a}{l_a}\stackrel{(2)}{=}\frac {\sum\, a\cos\frac {B-C}2}{2rs}\stackrel{(3)}{=}$ $\frac {\sum\, \left[2R\sin A\left(\cos\frac B2\cos\frac C2+\sin\frac B2\sin\frac C2\right)\right]}{2rs}\stackrel{(4)}{=}$

$\stackrel{(4)}{=}\frac {\sum\, \left[2R\sin A\left(\frac s{4R}\cdot\frac 1{\cos\frac A2}+\frac r{4R}\cdot\frac 1{\sin\frac A2}\right)\right]}{2rs}$ $\stackrel{(5)}{=}\frac {\sum\, \left[2\sin\frac A2\cos\frac A2\left(\frac s2\cdot\frac 1{\cos\frac A2}+\frac r2\cdot\frac 1{\sin\frac A2}\right)\right]}{2rs}=\boxed{\frac {s\sum\, \sin\frac A2+r\sum\, \cos\frac A2}{2rs}}$

where at steps $\left(1\right)\, ,\, \left(2\right)\, ,\, \left(3\right)\, ,\, \left(4\right)\, ,\, \left(5\right)$ I used the well known identities: $\left\| \begin{array}{cc} 1\blacktriangleright & 2S=ah_a=2rs\ \ ;\ \ 2\blacktriangleright\ \frac {h_a}{l_a}=\cos\frac {B-C}2 \\ \\ 3\blacktriangleright & \left\{\begin{array}{c} a=2R\sin A \\ \\ \cos\frac {B-C}2=\cos\frac B2\cos\frac C2+\sin\frac B2\sin\frac C2\end{array}\right. \\ \\ 4\blacktriangleright & \begin{array}{c} \prod\, \cos\frac A2=\frac s{4R}\ ;\ \prod\, \sin\frac A2=\frac r{4R}\end{array} \\ \\ 5\blacktriangleright &\, \ \sin A=2\sin\frac A2\cos\frac A2\end{array}\right\|$

In addition, since $\boxed{\prod\, \left(b+c\right)=2s\left(s^2+r^2+2Rr\right)}$ and $\boxed{\prod\, \tan\frac A2=\frac rs}$ the proposed inequality is now equivalent to :

$\frac {s\sum\, \sin\frac A2+r\sum\, \cos\frac A2}{2rs}\, \ge\, \frac {\sqrt{2s^2\left(s^2+r^2+2Rr\right)}}{4Rrs}\iff s$ $\sum\, \sin\frac A2+r\sum\, \cos\frac A2\, \ge\, s\cdot\sqrt{\frac {s^2+r^2+2Rr}{2R^2}}$

$\iff \sum\, \sin\frac A2+\prod\, \tan\frac A2\sum\, \cos\frac A2\, \ge\, \sqrt{\frac {s^2+r^2+2Rr}{2R^2}}$ .

But since in any $\triangle\, ABC$ the following strong inequality holds : $\boxed{\ \sum\, \cos\frac {B-C}2\, \ge\, 1+\sqrt{\frac {s^2+r^2+2Rr}{2R^2}}\ }$ (see here)

it suffices to prove that : $\boxed{\ 1+\sum\, \sin\frac A2+\prod\, \tan\frac A2\sum\, \cos\frac A2\, \ge\, \sum\, \cos\frac {B-C}2\ }\ \ \left(\ast\right)$ . By making the following

substitutions : $\left\{\begin{array}{lll} A & = & 180^{\circ}-2X \\ B & = & 180^{\circ}-2Y \\ C & = & 180^{\circ}-2Z\end{array}\right.$ where $\left\|\begin{array}{cccc} X,Y,Z\in\left(0,90^{\circ}\right) \\ \\ X+Y+Z=180^{\circ}\end{array}\right\|$ we will transform the inequality $\left(\ast\right)$ (in any triangle)

into one restricted to an acute-angled triangle : $\boxed{\ 1+\sum\, \cos X+\prod\, \cot X\sum\, \sin X\, \ge\, \sum\, \cos\left(Y-Z\right)\ }\ \ \left(\ast\ast\right)$ .

For convenience, we will use the notations $R$ , $r$ , $s$ for denoting the classical elements of $\triangle\, XYZ$ (which are now different from

those appearing above). Recall the identities : $\left\|\begin{array}{cccc} \sum\, \cos X=1+\frac rR\ \ ;\ \ \prod\, \cot X=\frac {s^2-\left(2R+r\right)^2}{2rs} \\ \\ \sum\, \sin X=\frac sR\ ;\ \sum\, \cos\left(Y-Z\right)=\frac {s^2+r^2+2Rr-2R^2}{2R^2} \end{array}\right\|$ in order to see

that the inequality $\left(\ast\ast\right)$ is equivalent to: $\frac {2R+r}R+\frac {s^2-\left(2R+r\right)^2}{2rs}\cdot\frac sR\ge\frac {s^2+r^2+2Rr-2R^2}{2R^2}$ which further becomes:

$2Rr\left(2R+r\right)+R\left[s^2-\left(2R+r\right)^2\right]\ge r\left(s^2+r^2+2Rr-2R^2\right)$ $\iff$ $\boxed{\ s^2\ge\frac {4R^3-2R^2r+Rr^2+r^3}{R-r}\ }\ (G)$

Finally, we will show that the last "boxed" inequality is true in any non-obtuse triangle. To do this, we'll make use of two well known

results, namely : $\boxed{\ s^2\ge 2R^2+10Rr-r^2-\sqrt{4R\left(R-2r\right)^3}\ }$ (Blundon) and $\boxed{\ s\ge 2R+r\ }$ , with the latter holding

in non-obtuse triangles. To be more specific, we will prove inequality $(G)$ in two stages: in the first one we show that the inequality $(G)$ holds for $R\le\left(1+\sqrt 2\right)r$ when compared to Blundon, while in the second stage we prove that $(G)$ is fulfilled for $R\ge\left(1+\sqrt 2\right)r$ when compared to $s\ge 2R+r$ . In the following lines we will use the substitution: $t=\frac Rr$ where $t\ge 2$ (Euler) in order to make the computations easier. Thus:
$\boxed{\ \begin{array}{cccc} \left(G_1\right) & 2R^2+10Rr-r^2-\sqrt{4R\left(R-2r\right)^3}\, \ge\, \frac {4R^3-2R^2r+Rr^2+r^3}{R-r} \\ \\ \iff & \left(t-1\right)\left(2t^2+10t-1-\sqrt{4t\left(t-2\right)^3}\, \right)\, \ge\, 4t^3-2t^2+t+1 \\ \\ \iff & \left(t-1\right)\left(2t^2+10t-1\right)-\left(4t^3-2t^2+t+1\right)\, \ge\, \left(t-1\right)\sqrt{4t\left(t-2\right)^3} \\ \\ \iff & t\left(t-2\right)\left(3-t\right)\, \ge\, \left(t-1\right)\sqrt{t\left(t-2\right)^3} \\ \\ \stackrel{\text{For}\ t\le 3}{\iff} & \left(t-2\right)^2\left(-t^2+2t+1\right)\, \ge\, 0 \\ \\ \implies & \left(G_1\right)\text{ holds for }\boxed{2\le t\le 1+\sqrt 2} \\ \\ \hline\hline \\\ \left(G_2\right) & \left(2R+r\right)^2\, \ge\, \frac {4R^3-2R^2r+Rr^2+r^3}{R-r} \\ \\ \iff & \left(t-1\right)\left(2t+1\right)^2\, \ge\, 4t^3-2t^2+t+1 \\ \\ \iff & t^2-2t-1\, \ge\, 0 \\ \\ \implies & \left(G_2\right)\text{ holds for }\boxed{t\ge 1+\sqrt 2}\end{array}\ }$ We conclude that the previous inequality $\left(G\right)$ holds true in any non-obtuse triangle, which means that $\left(\ast\ast\right)$ is also true in such triangles
so our proof is now complete. Equality occurs in the original inequality if and if $\triangle\, ABC$ is equilateral.

Remarks. As said by quykhtn-qa1, the proposed inequality represents a refinement of an old inequality (which also appeared in

the Canadian magazine CRUX MATHEMATICORUM long time ago - look for problem 2029*) : $\boxed{\ l_al_b+l_bl_c+l_cl_a\, \ge\, 3S\sqrt 3\ }$

I will now present a sketch of the aforementioned fact. Multiplying the initial inequality by $\boxed{\ \prod\, l_a=\tfrac {16Rr^2s^2}{s^2+r^2+2Rr}\ }$ one can easily

obtain the following equivalent enunciation i.e. $\boxed{\ l_al_b+l_bl_c+l_cl_a\, \ge\, \tfrac {4rs^2\sqrt 2}{\sqrt{s^2+r^2+2Rr}}\ }$ thus remaining to prove that :

$4s\sqrt 2\, \ge\, 3\sqrt{3\left(s^2+r^2+2Rr\right)}$ which simplifies to the following easy to prove inequality : $5s^2\, \ge\, 54Rr+27r^2$ .