Please, even more than the solution I would like to understand how get better at solving inequalities. Currently my only method is to just blindly try different manipulations to see if they work.
This problem comes from the book C-S Master Class so I think there is a solution by C-S, but others are welcome too. Here are some things I tried for this inequality:
Trial 1: Let $w = (x+y)(y+z)(x+z)$. Multiplying both sides by $w$ makes it
$$xw+yw+zw \le x^2(x+z)(x+y)+y^2(x+y)(y+z)+z^2(x+z)(y+z)$$
The LHS looks like C-S so I tried to apply C-S to the sequences $\{x(y+z), y(x+z), z(x+y)\}$ and $\{(x+y)(x+z), (x+y)(y+z), (x+z, y) \}$. So it remain to check that my RHS is equal the the RHS of the C-S inequality; but the degrees don't match.
Trial 2: Let $a = x+y, b=y+z, c = x+z$. Then the inequality becomes $$a+b+c \le \dfrac {(a-b+c)^2}{b}+ \dfrac {(b-c+a)^2}{c}+\dfrac {(c-a+b)^2}{a}$$
and from here I don't know where I would apply C-S.
By C-S $$\sum_{cyc}\frac{(a-b+c)^2}{b}\geq\frac{\left(\sum\limits_{cyc}(a-b+c)\right)^2}{\sum\limits_{cyc}b}=\frac{(a+b+c)^2}{a+b+c}=a+b+c.$$ The best way I think is C-S: $$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+z)}=\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}.$$ I see also at least five different proofs.