Prove that if $a,b,c≥0$, then $\sqrt[3]{abc}≤\frac{1}{3}(a+b+c)$ I'm just starting to learn how to do proofs and this was one of the exercises I'm trying to solve. A hint I have is:
The numbers $a,b,c$ can be expressed as $a=r^3$, $b=s^3$, $c=t^3$ for positive numbers $r,s,t$
The solution given from the book I'm reading is:
Observe that $r^3+s^3+t^3-3rst=\frac{1}{2}(r+s+t)[(r-s)^2+(s-t)^2+(t-r)^2]$
I can get to the part $r^3+s^3+t^3-3rst$ but there is no way I would have noticed that $r^3+s^3+t^3-3rst=\frac{1}{2}(r+s+t)[(r-s)^2+(s-t)^2+(t-r)^2]$. Is it okay to not be able to do this kind of proofs? The book I'm reading "Mathematical Proofs: A Transition to Advanced Mathematics" and pretty much any other book about proofs teaches you the basic proof techniques, but these techniques are just not enough to solve some exercises, because you need to "see" some crazy equivalence to be able to solve them. So my question is, should I strive to learn how to recognize these "crazy" equivalences or it's enough (for an average mathematics curriculum) to be able to solve the basic exercises? (with basic I mean exercises that don't require complex algebraic equivalences)
 A: I think this may be just a bad/non-instructive example given to you. Let this not put you off studying proofs in general and inequalities in particular.
The example is bad because it is asking you to do something without giving you any tools to do it. In fact, for many inequalities, the very thing you've been asked to prove (arithmetic-geometric mean inequality, or AM-GM for short) is a tool of choice. Namely, it can be proven that, for any finite number of positive real numbers $a_1, a_2,\ldots,a_n$, it is true that $\frac{1}{n}(a_1+a_2+\ldots+a_n)\ge\sqrt[n]{a_1a_2\cdots a_n}$ and that the equality is achieved if and only if all $a_i$'s are equal to each other.
This particular proof is instructive to a point but, again, not for a beginner. There is some value in trying to make it into a polynomial inequality, in this case by substitution; you could also try to cube the left side and get $(a+b+c)^3\ge 27abc$ (which would also work but will initially lead to a more complicated expression). The idea to try to complete the squares (such as $(r-s)^2$ etc.) is also sometimes helpful.
I guess the author of those exercises/book may have wanted to do a demonstration here rather than ask you to really prove it.
As for my advice for you: what I would suggest is to:

*

*Get used to some common inequalities (AM-GM is one of them, but there are many other "tools in the shed"), e.g. start with Wikipedia "inequalities" category. For the most important elementary ones, see below (apologies if I missed some):

*

*Cauchy-Schwarz inequality,

*Chebyshev's sum inequality,

*Generalized mean inequality,

*Jensen's inequality (requires knowledge of what is a convex function, but is extremely powerful),

*Muirhead's inequality (of which $r^3+s^3+t^3\ge 3rst$ is a very small special case),

*Rearrangement inequality etc.



*Browse MSE for tag inequality and see what questions/problems are there and what people are doing to solve them.
Once you are more familiar with the tools and how they are used, I guess you will start seeing patterns and be able to do some of those problems yourself.
A: In my opinion, they should at least have given a hint.
This is known as Gauß' identity. It is  a classic exercise in mid/high-school and there are actually two (equivalent) factorisations:
\begin{align}
r^3+s^3+t^3-3rst&=(r+s+t)(r^2+s^2+t^2-rs-st-tr)\\[1ex]
&=\frac{1}{2}(r+s+t)\bigl[(r-s)^2+(s-t)^2+(t-r)^2\bigr].
\end{align}
It can be easily be proved from the trinomial formula:
\begin{align}
(r+s+t)^3&=r^3+s^3+t^3+3(r^2s+s^2r+s^2t+st^2+t^2r+t^2)+6rst \\
&= r^3+s^3+t^3+3\bigl(rs(r+s)+st+st(s+t)+tr(r+t)\bigr)+6rst 
 \end{align}
We deduce a factorisation of
\begin{align}
&r^3+s^3+t^3-3rst =\\&=(r+s+t)^3-3\bigl(rs(r+s)+st+st(s+t)+tr(r+t)\bigr)-9rst \\
&= (r+s+t)^3-3\bigl(rs(r+s+t)
+st(r+s+t)+tr(r+s+t)\bigr)+9rst -9rst \\
&=(r+s+t)\bigl((r+s+t)^2-3rs-3st-3tr)\\
&=\cdots
 \end{align}
