Simple Inequality 3

Notice $(ab+bc+ca)\left(\dfrac{a}{2b+3c}+\dfrac{b}{2c+3a}+\dfrac{c}{2a+3b}\right)$ $=(ab+bc+ca)\left(\dfrac{a^2}{2ab+3ca}+\dfrac{b^2}{2bc+3ab}+\dfrac{c^2}{2ca+3bc}\right)$ By Titu's Lemma, this is greater than or equal to $\geq (ab+bc+ca)\left(\dfrac{(a+b+c)^2}{5ab+5bc+5ca}\right)=\dfrac{n^2}{5}$ If this is $\geq 20$ , then $n\geq 10$ and the minimum $n$ is $10$ .

Well, it might be easier than thought but I don't know, I made it difficult(my solution would seem to).

First of all, using Re-arrangement inequality,

$(\frac{a}{2b+3c} + \frac{b}{2c+3a} + \frac{c}{2a+3b}) \geq (\frac{a}{2c+3a} + \frac{b}{2a+3b} + \frac{c}{2b+3c})$

$\geq (\frac{a}{2b+3c} + \frac{b}{2c+3a} + \frac{c}{2a+3b})$

$\geq (\frac{a}{2c+3a} + \frac{b}{2b+3c} + \frac{c}{2a+3b})$

Adding all these,

$3(\frac{a}{2b+3c} + \frac{b}{2c+3a} + \frac{c}{2a+3b}) \geq (\frac{a+b+c}{2c+3a} + \frac{a+b+c}{2a+3b} + \frac{a+b+c}{2b+3c})$

$\geq (a+b+c)(\frac{1}{2c+3a} + \frac{1}{2b+3c} + \frac{1}{2a+3b})$

Now using Cauchy-Schwarz in Engel form,

$(\frac{1}{2c+3a} + \frac{1}{2b+3c} + \frac{1}{2a+3b}) \geq \frac{9}{5(a+b+c)}$

Now,

$3(\frac{a}{2b+3c} + \frac{b}{2c+3a} + \frac{c}{2a+3b}) \geq (a+b+c)(\frac{9}{5(a+b+c)})$ - (1)

Also,

$ab + bc + ca \geq \frac{{(a+b+c)}^{2}}{3}$ -(2)

Multiplying (1) and (2) and replacing $a + b + c$ with $n$

$(\frac{a}{2b+3c} + \frac{b}{2c+3a} + \frac{c}{2a+3b})(ab + bc + ca) \geq \frac{{n}^{2}}{5}$

$\frac{{n}^{2}}{5} \geq 20$

$n \geq 10$

Since the given expression is cyclic, we can assume that $a = b = c = \frac{n}{3}$ .

Substituting this value, the expression is reduced to :

$(\frac{n^2}{9} + \frac{n^2}{9} + \frac{n^2}{9})(\frac{1}{5} + \frac{1}{5} + \frac{1}{5}) \geq 20$

$=> \frac{n^2}{3} * \frac{3}{5} \geq 20$

$=> n^2 \geq 100$

$=> n \geq 10$

Thus the minimum value of $n$ is $10$ .

Just a comment. The rearrangement inequality cannot be applied here because 2c + 3a is not necessarily greatest than 2a+3b