Given three postive numbers $a, b, c$ so that $a\geqq b\geqq c$. Prove that $$\sum\limits_{cyc}\frac{a+ b\sqrt{\frac{b}{c}}}{a\sqrt{\frac{b}{c}}+ b}\geqq 3$$
I make it
Firstly, we need to have one general inequality $$\sum\limits_{cyc}\frac{a+ bW}{aW+ b}\geqq 3$$ By buffalo-way, let $c= 1, b= 1+ u, a= 1+ u+ v$ so $W= \sqrt{1+ u}$, be homogeneous. I guess that $${\rm W}= \sqrt{\frac{b}{c}}$$ An inspiration-one Given three positive numbers $a,b,c$ so that $a\leqq b\leqq c$. Prove that $\sum\limits_{cyc}\frac{a+1.4b}{1.4a+b}\geqq 3$ .
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, we need to prove that $$\sum_{cyc}\frac{a+\sqrt{\frac{b}{c}}b}{\sqrt{\frac{b}{c}}a+b}\geq3$$ or $$\sum_{cyc}\frac{x^2z+y^3}{x^2y+y^2z}\geq3.$$ Now, by AM-GM $$\sum_{cyc}\frac{x^2z+y^3}{x^2y+y^2z}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(y^3+x^2z)}{\prod\limits_{cyc}(x^2y+y^2z)}}$$ and it's enough to prove that $$\prod\limits_{cyc}(x^3+z^2y)\geq\prod\limits_{cyc}(x^2y+y^2z)$$ or $$\sum_{cyc}(x^5z^4+x^6y^2z)\geq xyz\sum_{cyc}(x^3y^3+x^4yz),$$ which is true by Rearrangement twice: $$\sum_{cyc}x^5z^4=x^4y^4z^4\sum_{cyc}\frac{x}{y^4}\geq x^4y^4z^4\sum_{cyc}\frac{x}{x^4}=xyz\sum_{cyc}x^3y^3$$ and $$\sum_{cyc}x^6y^2z=x^2y^2z^2\sum_{cyc}\frac{x^4}{z}\geq x^2y^2z^2\sum_{cyc}\frac{x^4}{x}=xyz\sum_{cyc}x^4yz.$$ Done!