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Prove that in a triangle $sumlimits_{cyc}frac{w_bw_c}{w_a}geqfrac{3}{4}left(sumlimits_{cyc}frac{a^2w_a}{w_bw_c}right)geqsqrt{3}s$

Mathematics Asked on December 15, 2021

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^2geq 0\
iff & sum x^3(y + z)(x – y)(x – z) + 11(x – y)^2(y – z)^2(z – x)^2 + 40sum y^2z^2(x – y)(x – z))+ 4xyzsum x(x – y)(x – z) + 9xyzsum (y + z)(x – y)(x – z) geq 0
end{aligned}

In this solution how step $$sum x^3(y + z)(x – y)(x – z) + 11(x – y)^2(y – z)^2(z – x)^2 + 40sum y^2z^2(x – y)(x – z))+ 4xyzsum x(x – y)(x – z) + 9xyzsum (y + z)(x – y)(x – z) geq 0$$comes from the step $$4w_b^2w_c^2 + 4w_c^2w_a^2 + 4w_a^2w_b^2 – 3a^2w_a^2 – 3b^2w_b^2 – 3c^2w_c^2geq 0$$

and also how to prove the right inequality

One Answer

The right inequality.

In the standard notation we need to prove that: $$sum_{cyc}frac{a^2cdotfrac{2bccosfrac{alpha}{2}}{b+c}}{frac{2accosfrac{beta}{2}}{a+c}cdotfrac{2abcosfrac{gamma}{2}}{a+b}}geqfrac{2(a+b+c)}{sqrt3}$$ or $$sum_{cyc}frac{a^2cdotfrac{2bcsqrt{frac{1+frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}}{frac{2acsqrt{frac{1+frac{a^2+c^2-b^2}{2ac}}{2}}}{a+c}cdotfrac{2absqrt{frac{1+frac{a^2+b^2-c^2}{2ab}}{2}}}{a+b}}geqfrac{2(a+b+c)}{sqrt3}$$ or $$sum_{cyc}frac{frac{a^2sqrt{bc(a+b+c)(b+c-a)}}{b+c}}{frac{sqrt{ac(a+b+c)(a+c-b)}}{a+c}cdotfrac{sqrt{ab(a+b+c)(a+b-c)}}{a+b}}geqfrac{2(a+b+c)}{sqrt3}$$ or $$sum_{cyc}frac{a(a+b)(a+c)}{b+c}sqrt{frac{b+c-a}{(a+b-c)(a+c-a)}}geqfrac{2sqrt{(a+b+c)^3}}{sqrt3}$$ or$$sum_{cyc}frac{a(a+b)(a+c)(b+c-a)}{b+c}geq2sqrt{frac{(a+b+c)^3prodlimits_{cyc}(a+b-c)}{3}}.$$ 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: $$sum_{cyc}frac{x(y+z)(2y+x+z)(2z+x+y)}{2x+y+z}geq8sqrt{frac{(x+y+z)^3xyz}{3}}.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.

Thus, we need to prove that: $$sum_{cyc}frac{(3v^2-yz)(3u+y)^2(3u+z)^2}{prodlimits_{cyc}(3u+x)}geq24sqrt{u^3w^3}$$ or $$frac{-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4}{w^3+54u^3+9uv^2}geq8sqrt{u^3w^3}$$ 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 $$ugeq vgeq w,$$ we obtain: $$f'(v^2)=405u^4-108u^2v^2-9uw^3-72u^2sqrt{uw^3}>0,$$ 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 $$frac{2x(x+3)^2}{2x+2}+frac{2(x+1)(x+3)(2x+2)}{x+3}geq8sqrt{frac{(x+2)^3x}{3}}$$ or $$x(x+3)^2+4(x+1)^3geq8(x+1)sqrt{frac{(x+2)^3x}{3}},$$ which after squaring of the both sides gives $$(x-1)^2(11x^4+50x^3+91x^2+88x+48)geq0$$ and we are done!

Answered by Michael Rozenberg on December 15, 2021

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