Find the value of $3a+b+c$, when $\frac{3a^2+b^2+3c^2}{ab+bc+ca}$ attains its minimum, where $abc=432$ and $(a,b,c)\in\mathbb R_{+}^{3}$

Question

Let $a,b,c$ be real positive numbers such that the expression $\dfrac{3a^2+b^2+3c^2}{ab+bc+ca}$ attains minimum possible value . If $abc=432$, then what value will be the expression $3a+b+c\thinspace \thinspace ?$

---

I think i have got a heads up to the question. For the expression to attain a minimum possible value $a<b<c$.Also the numbers should be seperated by a large magnitude(i speculate the min value of seperation to be 9). After that i am kinda stuck. Can anybody suggest anyother approach. Thanks.

Answer 1

Also, smoothing helps.

Indeed, let $k$ be a minimal value of the expression $\frac{3a^2+b^2+3c^2}{ab+ac+bc}$ and let $a+c=constant.$

Thus,$k>0$ and we see that it's enough to prove the inequality $$3(a+c)^2-6ac+b^2\geq k(ac+b(a+c))$$ for a maximal value of $ac$, which happens for $a=c$.

Thus, we need to find a maximal value of $k$ for which the inequality $$6a^2+b^2\geq k(b^2+2ab)$$ or $$(6-k)a^2+b^2\geq2kab$$ is true for any positives $a$ and $b$.

Since should be $6-k>0,$ by AM-GM we obtain: $$(6-k)a^2+b^2\geq2\sqrt{6-k}ab,$$ which gives that should be $$2\sqrt{6-k}ab\geq2kab,$$ or $$k^2+k-6\leq0$$ or $$k\leq2.$$

The equality occurs for $a=c$ and $b=2c$, which says that $2$ is a minimal value and $a^2\cdot2a=432,$ which gives $(a,b,c)=(6,12,6)$ and $$3a+b+c=18+12+6=36.$$

Answer 2

$\rm\bf{Task}$

Let $a,b,c$ be real positive numbers such that the expression $\dfrac{3a^2+b^2+3c^2}{ab+bc+ca}$ attains minimum possible value . If $abc=432$, then what value will be the expression $3a+b+c\thinspace \thinspace ?$

---

Using some substitutions will work for us .

Substituting $\thinspace \dfrac ab=m,\thinspace \dfrac cb=n\thinspace $, then we have :

$$\frac{3m^2+1+3n^2}{m+n+mn}$$

Without loss of generality, we assume that, $\thinspace m≥n>0\thinspace $ .

Then, again substituting $\thinspace m=r+s,\thinspace n=r-s\thinspace$, where $\thinspace r>s≥0\thinspace $, we obtain :

$$ \begin{align}\frac {6(r^2+s^2)+1}{2r+r^2-s^2}&≥\frac {6r^2+1}{r^2+2r}=p\end{align} $$

This leads to the following :

$$ \begin{align}&\left(p-6\right)r^2+2pr-1=0\\ &\Delta_r=p^2+p-6≥0\\ \implies &(p-2)(p+3)≥0\\ \implies &p≥2\end{align} $$

Thus, if $\thinspace p=2\thinspace$, then $\thinspace r=-\dfrac {p}{p-6}=\dfrac 12\thinspace $ .

or equivalently :

$$ \begin{align}\begin{cases}s=0\\ r=\dfrac 12\end{cases} \implies \begin{cases}a=c\\ \dfrac ab=\dfrac 12\end{cases}\end{align} $$

which implies :

$$ \begin{align}&abc=a\cdot2a\cdot a=432\\ \implies &a=6,\thinspace b=12,\thinspace c=6\\ \implies &3a+b+c=36\end{align} $$

You are done .

Answer 3

Using $ac \le (a + c)^2/4$, we have \begin{align*} \frac{3a^2+b^2+3c^2}{ab+bc+ca} &= \frac{3a^2+b^2+3c^2}{b(c + a) + ca}\\ &\ge \frac{3a^2+b^2+3c^2}{b(c + a) + (c+a)^2/4}\\ &= \frac{3(a+c)^2 - 6ac + b^2}{b(c + a) + (c+a)^2/4}\\ &\ge \frac{3(a+c)^2 - 6(a+c)^2/4 + b^2}{b(c + a) + (c+a)^2/4}\\ &= -\frac12 + \frac{4b + a + c}{4(a+c)} + \frac{25(a+c)}{4(4b + a + c)}\\ &\ge -\frac12 + 2\sqrt{ \frac{4b + a + c}{4(a+c)} \cdot \frac{25(a+c)}{4(4b + a + c)}}\\ &= 2 \end{align*} with equality if $a = c$ and $b = 2c$, where we do a partial fraction decomposition for $b$ in the middle, and use AM-GM in the last 2nd step.
(Note: Once we motivate the minimum $2$, we can write a shorter solution: $\frac{3a^2+b^2+3c^2}{ab+bc+ca}-2 = \frac{(b-a-c)^2 + 2(a-c)^2}{ab + bc + ca} \ge 0$. So the minimum is $2$ when $a = c$ and $b = 2c$.)

Using $abc = 432$, we have $a = c = 6, b = 12$. Thus, $3a + b + c = 36$.