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.
Best Answer
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.