Question Number 198903 by necx122 last updated on 25/Oct/23
$${Find}\:{the}\:{minimum}\:{value}\:{of}\: \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\:{for}\:{all}\:{positive}\:{real} \\ $$$${numbers} \\ $$
Commented by necx122 last updated on 25/Oct/23
please I need help with this
Answered by AST last updated on 25/Oct/23
![=(a^2 /(ab+ac))+(b^2 /(bc+ab))+(c^2 /(ac+bc)) ≥(((a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca))/(2(ab+bc+ca)))≥((3(ab+bc+ca))/(2(ab+bc+ca))) =(3/2)[Equality when a=b=c]](https://www.tinkutara.com/question/Q198916.png)
$$=\frac{{a}^{\mathrm{2}} }{{ab}+{ac}}+\frac{{b}^{\mathrm{2}} }{{bc}+{ab}}+\frac{{c}^{\mathrm{2}} }{{ac}+{bc}} \\ $$$$\geqslant\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\mathrm{2}\left({ab}+{bc}+{ca}\right)}\geqslant\frac{\mathrm{3}\left({ab}+{bc}+{ca}\right)}{\mathrm{2}\left({ab}+{bc}+{ca}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left[{Equality}\:{when}\:{a}={b}={c}\right] \\ $$
Commented by necx122 last updated on 25/Oct/23
wow!!! This is a great solution sir. Please, I'm quite new to this manner of questions. Sir AST, can you please suggested sources where I can learn this properly. Thank you.
Commented by AST last updated on 25/Oct/23
$${This}\:{is}\:{just}\:{an}\:{application}\:{of}\:{Cauchy}−{Schwarz} \\ $$$$\left({Engel}'{s}\:{Form}\right). \\ $$
Commented by necx122 last updated on 25/Oct/23
wow! This sounds like a sort of advanced level maths but how come I'm seeing such a question in junior mathematics olympiad? Could it mean that students in the junior school are really expected to know this?
It's indeed overwhelming.
Commented by AST last updated on 25/Oct/23
$${Not}\:{really}\:{advanced},{they}\:{are}\:{standard}\:{techniques} \\ $$$${in}\:“{olympiad}''\:{competitions}. \\ $$
Commented by necx122 last updated on 25/Oct/23
hmmm... This is indeed a lot. Thank you for this information sir. I'm grateful.