Finding $\small{\min\limits_{ab+bc+ca=3}\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}.}$

Question

Let $a,b,c> 0 :ab+bc+ca=3.$ Find the minimum$$P=\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}$$I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition. It is reasonable to set $a=b=c=1$ and we get minimum is $\dfrac{3}{5}.$ Now, we need to prove $$\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}\ge \frac{3}{5}$$It is equivalent to $$\frac{16\sum bc+4\sum a^2(b+c)+abc(a+b+c)}{(4a+bc)(4b+ca)(4c+ab)}\ge \frac{3}{5}$$I think we can use $uvw$ for the remain but it is not well -known method and the contestant need to prove it.

I hope to see more ideas. Thank you for interest.

Answer 1

Proof.

By using AM-GM$$\dfrac{1}{4a+bc} \ge \dfrac{1}{\dfrac{4a}{b+c}+a(b+c)+bc}=\dfrac{1}{\dfrac{4a}{b+c}+3}=\dfrac{b+c}{4a+3b+3c}.$$ (See also Nguyen Thai An's idea)

Similarly, it implies that we need to prove$$\dfrac{b+c}{4a+3(b+c)}+\dfrac{c+a}{4b+3(c+a)}+\dfrac{a+b}{4c+3(a+b)} \ge \dfrac{3}{5}.$$ The inequality is true by Cauchy-Schwarz.

Can you go futher

Indeed,\begin{align*} \sum\limits_{cyc}\dfrac{b+c}{4a+3(b+c)}&=\sum\limits_{cyc}\left(\dfrac{4(a+b+c)}{3(a+b+c)+a}-1\right)\\ &\ge 4(a+b+c)\cdot\dfrac{9}{10(a+b+c)}-3\\ &=\dfrac{3}{5}. \end{align*} Hence, we get desired minimal value.

Answer 2

Remarks: I am not sure if this is nice for contest. So I don't prove (1). Anyway, I just give some thoughts.

Some thoughts.

We use isolated fudging.

It suffices to prove that $$\frac{1}{4a + bc} \ge \frac{-3a + 24b + 24c}{75(a + b + c)}. \tag{1}$$ (Note: Taking cyclic sum on (1), we get $\frac{1}{4a+bc}+\frac{1}{4b+ca}+\frac{1}{4c+ab} \ge \frac35$.)

(1) is true which is verified by Mathematica.

Answer 3

For $a=b=c=1$ we obtain a value $\frac{3}{5}.$

We'll prove that it's a minimal value.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$.

Thus, $v=1$ and we need to prove that $$\sum_{cyc}\frac{v^2}{4va+bc}\geq\frac{3}{5}$$ or $$\sum_{cyc}v^2(4va+bc)(4vb+ac)\geq\frac{3}{5}\prod_{cyc}(4va+bc)$$ or $$v^2\sum_{cyc}(16v^2ab+4va^2b+4va^2c+a^2bc)\geq$$ $$\geq\frac{3}{5}\left(64v^3abc+a^2b^2c^2+16v^2\sum_{cyc}a^2b^2+4v\sum_{cyc}a^3bc\right)$$ or $$v^2(48v^4+4v(9uv^2-3w^3)+3uw^3)\geq$$ $$\geq\frac{3}{5}(64v^2w^3+w^6+16v^2(9v^4-6uw^3)+4vw^3(9u^2-6v^2)).$$

Now it's obvious that our inequality is equivalent to $f(w^3)\geq0,$ where $f$ is a concave function.

But the concave function gets a minimal value for an extremal value of $w^3$, which by $uvw$ happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$.

Thus, by AM-GM: $$LSH\rightarrow \frac{1}{4a}+\frac{1}{4b}+\frac{1}{ab}=\frac{a+b}{12}+\frac{1}{3}\geq\frac{2\sqrt3}{12}+\frac{1}{3}>\frac{3}{5}.$$

2. Two variables are equal.

Let $b=a$ and $c=\frac{3-a^2}{2a},$ where $0<a<\sqrt3$ and after this substitution we obtain: $$(a-1)^2(2-a)(11+6a-a^2)\geq0$$ and we are done!