<s>[INST] Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet  \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}{3a^2+ab}+\dfrac{b^2}{3b^2+bc}+\dfrac{c^2}{3c^2+ca}\ge\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)+ab+bc+ca}=\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}$
$\bullet$ $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}$ so we have $\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}\le \dfrac{3}{4}$ but we need to prove $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} \ge \dfrac{3}{4}$
Can you give me some hint ?
 [/INST] The value $\frac 3 4$ is a maximum, indeed we have that by $x=\frac b a$, $y=\frac cb$, $z=\frac a c$ with $xyz=1$
$$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} = \dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3} \le \frac 3 4$$
and
$$\frac34-\left(\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}\right) =\frac{3 x y z+ 5 y z+ 5 x z+ 5 x y+ 3 x + 3 y  + 3 z-27}{4 (3 + x) (3 + y) (3 + z)}\ge 0$$
since by AM-GM
$$3 x y z+5 y z+ 5 x z+ 5 x y+ 3 x + 3 y  + 3 z \ge 27\sqrt[27]{(xyz)^{18}}=27$$
with equality for $x=y=z=1$.
As noticed the expression has not a minimum, indeed by

*

*$b=ka \implies x=k$

*$c=kb \implies y=k$

*$z=\frac1 {xy}=\frac1{k^2}$
we obtain
$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}=\dfrac{1}{k+3}+\dfrac{1}{k+3}+\dfrac{k^2}{1+3k^2} \to \frac13$$
as $k \to \infty$ and
$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}-\frac13=\frac{53 + 9 (x+y+z)}{3 (x + 3) (y + 3) (z + 3)}>0$$
which shows that the value $\frac13$ is an infimum for the expression that is

$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}>\frac13$$

</s>