Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$ Let $ ABC$ is a triangle, $ w_a, w_b, w_c$ are bisectors of angles, $ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle, prove that:$$ \frac {w_bw_c}{w_a} + \frac {w_cw_a}{w_b} + \frac {w_aw_b}{w_c} \geq \frac {3}{4}\left(\frac {a^2w_a}{w_bw_c} + \frac {b^2w_b}{w_cw_a} + \frac {c^2w_c}{w_aw_b}\right) \geq \sqrt {3}s$$
I found the solution of this inequality
\begin{aligned}
& \frac {w_bw_c}{w_a} + \frac {w_cw_a}{w_b} + \frac {w_aw_b}{w_c} \geq \frac {3}{4}\left(\frac {a^2w_a}{w_bw_c} + \frac {b^2w_b}{w_cw_a} + \frac {c^2w_c}{w_aw_b}\right)\\
\iff & 4w_b^2w_c^2 + 4w_c^2w_a^2 + 4w_a^2w_b^2 - 3a^2w_a^2 - 3b^2w_b^2 - 3c^2w_c^2\geq 0\\
\iff & \sum x^3(y + z)(x - y)(x - z) + 11(x - y)^2(y - z)^2(z - x)^2 + 40\sum y^2z^2(x - y)(x - z))+ 4xyz\sum x(x - y)(x - z) + 9xyz\sum (y + z)(x - y)(x - z) \geq 0
\end{aligned}
In this solution how step $$\sum x^3(y + z)(x - y)(x - z) + 11(x - y)^2(y - z)^2(z - x)^2 + 40\sum y^2z^2(x - y)(x - z))+ 4xyz\sum x(x - y)(x - z) + 9xyz\sum (y + z)(x - y)(x - z) \geq 0$$comes from the step $$4w_b^2w_c^2 + 4w_c^2w_a^2 + 4w_a^2w_b^2 - 3a^2w_a^2 - 3b^2w_b^2 - 3c^2w_c^2\geq 0$$
and also how to prove the right inequality
 A: The right inequality.
In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}}{\frac{2ac\cos\frac{\beta}{2}}{a+c}\cdot\frac{2ab\cos\frac{\gamma}{2}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}}{\frac{2ac\sqrt{\frac{1+\frac{a^2+c^2-b^2}{2ac}}{2}}}{a+c}\cdot\frac{2ab\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{\frac{a^2\sqrt{bc(a+b+c)(b+c-a)}}{b+c}}{\frac{\sqrt{ac(a+b+c)(a+c-b)}}{a+c}\cdot\frac{\sqrt{ab(a+b+c)(a+b-c)}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{a(a+b)(a+c)}{b+c}\sqrt{\frac{b+c-a}{(a+b-c)(a+c-a)}}\geq\frac{2\sqrt{(a+b+c)^3}}{\sqrt3}$$ or$$\sum_{cyc}\frac{a(a+b)(a+c)(b+c-a)}{b+c}\geq2\sqrt{\frac{(a+b+c)^3\prod\limits_{cyc}(a+b-c)}{3}}.$$
Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sum_{cyc}\frac{x(y+z)(2y+x+z)(2z+x+y)}{2x+y+z}\geq8\sqrt{\frac{(x+y+z)^3xyz}{3}}.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$,  and $xyz=w^3$.
Thus, we need to prove that:
$$\sum_{cyc}\frac{(3v^2-yz)(3u+y)^2(3u+z)^2}{\prod\limits_{cyc}(3u+x)}\geq24\sqrt{u^3w^3}$$ or
$$\frac{-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4}{w^3+54u^3+9uv^2}\geq8\sqrt{u^3w^3}$$ or $f(v^2)\geq0,$ where
$$f(v^2)=-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4-8(w^3+54u^3+9uv^2)\sqrt{u^3w^3}.$$
But since by Maclaurin $$u\geq v\geq w,$$ we obtain:
$$f'(v^2)=405u^4-108u^2v^2-9uw^3-72u^2\sqrt{uw^3}>0,$$ which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous, it's enough to assume that $y=z=1,$ which gives
$$\frac{2x(x+3)^2}{2x+2}+\frac{2(x+1)(x+3)(2x+2)}{x+3}\geq8\sqrt{\frac{(x+2)^3x}{3}}$$ or
$$x(x+3)^2+4(x+1)^3\geq8(x+1)\sqrt{\frac{(x+2)^3x}{3}},$$ which after squaring of the both sides gives
$$(x-1)^2(11x^4+50x^3+91x^2+88x+48)\geq0$$ and we are done!
