How do you come up with great and ingenious questions In mathematics how are the professionals and great authors able to come up with such ingenious questions which which are so difficult yet elegant.
Like in integral calculus it involve a substitution or adding and subtracting sometimes or a trick which would make you say WOW! and you cannot understand like how anyone could come up with that. Like this or this or this
Or questions on co-ordinate geometry when they ask about making a tangent or something and then, something and to prove that it's always true for some conditions like how could you come up with that.
I am posing this question to understand what goes on in the minds of a paper setter who tries to make a difficult question how does he come up with new ideas how is it that he can come off the new and ingenious your difficult questions every time be it trigonometry,calculus,algebra and geometry
 A: I can show, how it can work in creating of inequalities.
We know that the $uvw$ method  (see here: https://artofproblemsolving.com/community/c6h278791 or here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mo&paperid=194&option_lang=rus ) is very useful. 
Sometimes this method gives a proof without great effort, but I want that my students will a bit of think.
We know that if our inequality is equivalent to $f\geq0,$ where $f$ is monotone function of $w^3$ or $f$ is a concave function of $w^3$, so we can use the $uvw$.
All this we can say if $f$ is a function of $u$ or of $v^2$.
We want to create an inequality, which very easy to write in the $uvw$'s language, but the method does not work or gives a very hard and ugly solution. 
For which we can take an easy expression, where $u$, $v^2$ and $w^3$ are involved and fixture it.
Id est, we got a condition in the inequality, which we want to create. 
We know that
$$(a+b+c)^3=\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=(a^3+b^3+c^3)+3(a+b)(a+c)(b+c).$$
Let $(a+b)(a+c)(b+c)$ be this expression, where $a$, $b$ and $c$ are positives.
If $(a+b)(a+c)(b+c)=8$, so
$$(a+b+c)^3=a^3+b^3+c^3+24=a^3+b^3+c^3+8\cdot3.$$ 
It's obvious that $a^3+b^3+c^3\geq3,$ but in the sum $a^3+b^3+c^3+8\cdot3$
the expression $a^3+b^3+c^3$  does not play a major role, 
which says that after using AM-GM we'll get a very strong inequality:
$$(a+b+c)^3=a^3+b^3+c^3+8\cdot3\geq9\sqrt[9]{(a^3+b^3+c^3)\cdot3^8},$$ which gives
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}.$$
Since $(a+b)(a+c)(b+c)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3,$ we see that our inequality is equivalent to $$u\geq\sqrt[27]{9u^3-9uv^2+w^3}$$ or $F(u,v^2,w^3)\geq0,$ where
$$F(u,v^2,w^3)=u^{27}-(9u^3-9uv^2+w^3)\left(\frac{9uv^2-w^3}{8}\right)^8.$$
We see that $uvw$ gives a very hard solution, but we got a  nice inequality, which we can prove in one line! 
