Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$

Question

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 ?

Answer 1

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

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

Answer 2

For $c\rightarrow0^+$ and $b\rightarrow+\infty$ we see that our expression is closed to $\frac{1}{3}$.

But $\sum\limits_{cyc}\frac{a}{b+3a}\geq\frac{1}{3}$ it's just $$\sum_{cyc}\left(9a^2b+\frac{53}{3}abc\right)\geq0,$$ which is obvious.

Answer 3

Another way.

We know that the expression is closed to $\frac{1}{3}$, but by C-S $$\sum_{cyc}\frac{a}{b+3a}=\frac{a^2}{ab+3a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+ab)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+6ab)}=\frac{1}{3},$$ which says that $\frac{1}{3}$ is an infimum.

Answer 4

Another way:

We know that our expression is closed to $\frac{1}{3}$ but $$\sum_{cyc}\frac{a}{b+3a}\geq \sum_{cyc} \frac{a}{3b+3a+3c}=\frac{1}{3}$$ and we got the same result again.