Seeking maximum parameter value so that inequality always holds true

Question

What is the largest positive integer $n$ such that $$\frac{a^2}{\frac{b}{29}+\frac{c}{31}} + \frac{b^2}{\frac{c}{29}+\frac{a}{31}} + \frac{c^2}{\frac{a}{29}+\frac{b}{31}} \;\ge\; n\,(a+b+c)$$ holds true for all positive real numbers $a,b,c\,$?

I tried using AM $\ge$ GM $\ge$ HM inequality on the numbers $\frac{1}{\frac{1}{a^2}\left(\frac{b}{29}+\frac{c}{31}\right)}, \frac{1}{\frac{1}{b^2}\left(\frac{c}{29}+\frac{a}{31}\right)}, \frac{1}{\frac{1}{c^2}\left(\frac{a}{29}+\frac{b}{31}\right)}$ since that seems like the only thing that comes to my mind when seeing a homogenous inequality, but that doesn't work at all.

Any help is appreciated.

Answer 1

AM $\geqslant$ HM is viable in its weighted form $\,\sum_k w_kx_k\geqslant\big(\sum_k w_k\,x^{-1}_k\big)^{-1}\,$: $$\frac a{a+b+c}\cdot\frac a{31b+29c} + \frac b{a+b+c}\cdot\frac b{31c+29a} + \frac c{a+b+c}\cdot\frac c{31a+29b}\\[3ex] \geqslant \left(\frac a{a+b+c}\cdot\frac {31b+29c}a + \frac b{a+b+c}\cdot\frac{31c+29a}b + \frac c{a+b+c}\cdot\frac{31a+29b}c\right)^{-1} = \frac 1 {60}$$ Hence the optimal parameter is $\frac{29\cdot 31}{60}\lessapprox 15$, therefore the largest positive integer is $\,n=14\,$.

Answer 2

Another solution relies on the Cauchy-Bunyakowski-Schwarz inequality in the form $$\begin{align} \sum_k\frac{x_k^2}{y_k} & \;\geqslant\;\frac{\left(\sum_k x_k\right)^2}{\sum_k y_k}\\[1.67ex] \implies\;\frac{a^2}{31b+29c} + \frac{b^2}{31c +29a} + \frac{c^2}{31a +29b} & \;\geqslant\; \frac{(a+b+c)^2}{31b+29c + 31c+29a + 31a +29b}\\[1.5ex] & \;=\;\frac1 {60}\,(a+b+c)\end{align}$$