<s>[INST] How to approach Inequalities that can be solved using Rearrangement Inequality? I am learning Inequalities and today I've encountered the famous Rearrangement Inequality. I'll state the theorem: 

Consider any two collections of real numbers in increasing order, $$a_1\leq a_2\leq....a_n$$ and $$b_1\leq b_2\leq....b_n$$ Then for any permutation ($a_1',a_2',....a_n'$) of ($a_1,a_2....,a_n$) it happens that: $$\sum_{i=1}^{n}a_ib_i\geq\sum_{i=1}^{n}a_i'b_i\geq\sum_{i=1}^{n}a_{n-i+1}b_i$$

Now, I understand what the theorem is saying, but I am finding it difficult to apply this theorem while solving problems. So I would like to know the thought process involved in solving these problems. In particular,
1) How does one identify inequalities that can be easily solved by the Rearrangement Inequality?  
2)What are some standard ways in which one applies the Rearrangement Inequality? For example, we may prove the given inequality using a direct application of RI or we may use two inequalities (derived using RI) and then prove the result.
I've  listed a few problems, for starters: 

 [/INST] 1.61 
$(a,b,c)$ and $(a^2,b^2,c^2)$ are the same ordered.
Thus, by Rearrangement $\sum\limits_{cyc}(a^2\cdot a)\geq\sum\limits_{cyc}a^2\cdot b$, which gives 
$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$
In another hand, $(ab,ac,bc)$ and $\left(\frac{1}{c^2},\frac{1}{b^2},\frac{1}{a^2}\right)$ are the same ordered.
Thus, by Rearrangement $\sum\limits_{cyc}\left(ab\cdot\frac{1}{c^2}\right)\geq\sum\limits_{cyc}\left(ab\cdot\frac{1}{a^2}\right)=\sum\limits_{cyc}\frac{a}{c}$ or
$$a^3b^3+a^3c^3+b^3c^3\geq a^2b+b^2c+c^2a$$
and we are done! 
1.62.
Let $a^2c=x$, $b^2a=y$ and $c^2b=z$.
Hence, we need to prove that $x^2+y^2+z^2\geq xy+xz+yz$,
which is Rearrangement because $(x,y,z)$ and $(x,y,z)$ are the same ordered.
1.63.
$\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are the same ordered.
1.64.
$(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered.
Thus, $2\sum\limits_{cyc}\frac{a}{b+c-a}\geq\sum\limits_{cyc}\left(\frac{b}{b+c-a}+\frac{c}{b+c-a}\right)=\sum\limits_{cyc}\frac{b+c}{b+c-a}$, which gives
$\sum\limits_{cyc}\frac{a}{b+c-a}\geq\sum\limits_{cyc}\frac{b+c-a}{b+c-a}=3$.
Done!
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