prove this inequality with $a+b+c=1$

Question

Let $a,b,c>0,a+b+c=1$,show that $$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge \dfrac{16}{3(a+b)(b+c)(c+a)}$$

Answer 1

This inequality I think some time,Now I solve it.Following is my solution.

Use Holder inequality we have $$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\cdot\sum_{cyc}c(a+b)^2\ge (a+b+b+c+c+a)^3$$ so $$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge\dfrac{8(a+b+c)^3}{\displaystyle\sum_{cyc}c(a+b)^2}=\dfrac{8}{6abc+\displaystyle\sum_{cyc}c(a^2+b^2)}$$ it suffices to show that $$\dfrac{8}{6abc+\displaystyle\sum_{cyc}c(a^2+b^2)}\ge\dfrac{16}{3(a+b)(b+c)(c+a)}$$ since $$(a+b)(b+c)(c+a)=2abc+\sum_{cyc}c(a^2+b^2)$$ it suffices to show $$\sum_{cyc}c(a^2+b^2)\ge 6abc$$ it is clear AM-GM inequality By Done