Prove that $\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) $ 
Let $x, y, z\in [a,b]$ and $u, v, w\in [a, b] $, where $0 <a <b $,  s.t. $x^2+y^2+z^2=u^2+v^2+w^2$. Show that $$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) .$$

I tried to apply Cauchy- Schwartz and the fact that $ \frac {a^4+b^4}{ab (a^2+b^2)} \geq \frac {a}{b} $ but I didn't succeed.
 A: The inequality is homogeneous, i.e. multiplying all variables with the same factor will not change it. This allows us to demand that the multiplied variables (for which we use the same variable letters as before) satisfy $x^2 + y^2 + z^2 = u^2 + v^2 + w^2 = 3$  and in turn  $a \le 1$ and $\sqrt{3-a^2} \ge b \ge 1$. So we have to prove 
$$
\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
By re-ordering, w.l.o.g., let $x \le y\le z$.
By the rearrangement inequality, the LHS gets as large as possible if we use, for $z^3$, the smallest possible $w$, which is $a$. Considering the choices of $x,y,z$ the highest value on the LHS will then be obtained for $z=b$.  So we have to prove
$$
\frac {x^3}{u}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Again by rearrangement, the converse is true for the term $\frac {x^3}{u}$, so we need to prove 
$$
\frac {a^3}{b}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now the remaining values are determined: $y^2=v^2=3-a^2-b^2$. So finally we need to show 
$$
\frac {a^3}{b}+(3-a^2-b^2) + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}
$$
which is 
$$
(a^2+b^2)(a^4 + ab(3-a^2-b^2) + b^4)    \leq 3 (a^4+b^4)
$$
Writing $3=a^2+b^2+v^2$ we have
$$
(a^2+b^2)(a^4 + ab v^2 + b^4)    \leq (a^2+b^2+v^2) (a^4+b^4)\\
\rightarrow \quad 0 \leq v^2 (b^3-a^3)(b-a)
$$
which is true. This proves the claim.
Note that for more than three terms, a similar inequality holds. See here.
