Note that an equality condition is possible. Condider $n$ even. Let half of the $x_i$ equal to $a$ and let the corresponding $y_i$ equal to $b$. For the other half of the variables, exchange $a$ and $b$. Then the condition $x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2 = \frac{n}2 (a^2 + b^2)$ holds. Inserting in the inequality, indeed we have
$$
\frac{n}2 (\frac{a^3}{b} + \frac{b^3}{a}) = \frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n} = \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) = \frac {a^4+b^4}{ab (a^2+b^2)} \frac{n}2 (a^2 + b^2)
$$
with equality.
Now let us show that this constellation is extremal.
Let $q_i = \frac{x_i}{y_i}$. Choose a set of $x_i$. W.l.o.g., order the $x_i^2$ in ascending order. We have for $s$, the sum of squares: $n a^2 \leq s = x_1^2+x_2^2+\ldots+x_n^2 =y_1^2+y_2^2+\ldots+y_n^2 \leq n b^2$
The inequality is
$$
(x_1^2 q_1+x_2^2 q_2+\ldots+x_n^2 q_n) \leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)
$$
Now we argue for some extremal set of $y_i$.
By the rearrangement inequality, the maximum value of the LHS is obtained if the highest factors $q_i$ are chosen for the highest $x_i^2$. This can be achieved by letting as many $y_i = a$ for the highest indices $i$. The number $k$ of these settings $y_i = a$ is limited by $s =y_1^2+y_2^2+\ldots+y_{n-k}^2 + k a^2$. The number $k$ can be made as large as possible if all of the previous $y_1 = \ldots = y_{n-k} = b$. Then we have $s =(n-k) b^2 + k a^2$. This is $k = \frac{nb^2 -s}{b^2 - a^2}$ or the nearest possible smaller integer. So we have to show
$$
\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n} \leq \frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a} \leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)
$$
or
$$
\frac{\frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a}}{x_1^2+x_2^2+\ldots+x_n^2} \leq \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now again, for the LHS the maximum value has to be found by considering settings of the $x_i$. This maximal setting of the $x_i$ is obtained if most of the $x_i$ with high indices $i$ are chosen as high as possible, i.e. $x_i=b$. With the same argument as already given, this is possible for $n-k$ many $b$. This gives that we have to show (for $k \leq n/2$):
$$
\frac{\frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a}}{x_1^2+x_2^2+\ldots+x_n^2} \leq
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2} \leq
\frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now in the LHS, we have that
$$
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2}
$$
is $1$ for $k=0$ and maximal for $k = n/2$ where we have
$$
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2} \quad {{(k = n/2)}\atop {=}} \quad
\frac {a^4+b^4}{ab (a^2+b^2)}
$$
The same method can be applied for $k \geq n/2$ where again the LHS is $1$ for $k=n$.
For odd $n$ we have to take extra care of the central term with $i = (n+1)/2$.
This proves the claim.
One can proceed similarly as in the proof of the “regular” rearrangement inequality: If $\sigma$ is a permutation of $\{1, \ldots ,n\}$ and not the identity then there are indices $j < k$ such that exchanging $\sigma(j)$ and $\sigma(k)$ gives a new permutation $\tau$ with more fixed points than $\sigma$ and
$$ \tag{*}
\sum_{i=1}^n f(x_i + y_{\sigma(i)}) \le \sum_{i=1}^n f(x_i + y_{\tau(i)}) \, .
$$
If $\tau$ is not the identity then this step can be repeated, and after finitely many steps one obtains
$$
\sum_{i=1}^n f(x_i + y_{\sigma(i)}) \le \sum_{i=1}^n f(x_i + y_i) \, .
$$
In the case of the “regular” rearrangement inequality one uses that for $a_1 \le a_2$ and $b_1 \le b_2$
$$
(a_2-a_1)(b_2-b_1) \ge 0 \implies a_1 b_2 + a_2 b_1 \le a_1 b_1 + a_2 b_2 \, .
$$
In our case one can use the following to prove $(*)$:
If $f$ is a convex function and $a_1 \le a_2$ and $b_1 \le b_2$ then
$$
f(a_1 + b_2) + f(a_2 + b_1) \le f(a_1 + b_1) + f(a_2 + b_2) \, .
$$
This holds trivially if $a_1 =a_2$ or $b_1 = b_2$. In the case $a_1 < a_2$ and $b_1 < b_2$ it follows from adding the convexity conditions:
$$
f(a_1 + b_2) \le \frac{a_2-a_1}{a_2+b_2-a_1-b_1} f(a_1 + b_1) + \frac{b_2 - b_1}{a_2+b_2-a_1-b_1} f(a_2 + b_2) \\
f(a_2 + b_1) \le \frac{b_2-b_1}{a_2+b_2-a_1-b_1} f(a_1 + b_1) + \frac{a_2 - a_1}{a_2+b_2-a_1-b_1} f(a_2 + b_2)
$$
For positive sequences $u_1, \ldots, u_n$ and $v_1, \ldots, v_n$ the normal rearrangement inequality follows from the generalized one with $f(t)=e^t$ applied to $x_i = \log u_i$ and $y_i = \log v_i$, since then
$$
f(x_i + y_{\sigma(i)}) = u_i \cdot v_{\sigma(i)} \ .
$$
It is also a consequence of Karamata's inequality: Set
$$
(a_1, a_2, \ldots , a_n) = (x_n + y_n, x_{n-1}+y_{n-1}, \ldots, x_1 + y_1)
$$
and let $(b_1, b_2, \ldots , b_n)$ be a decreasing rearrangement of
$$
(x_n + u_n, x_{n-1}+u_{n-1}, \ldots, x_1 + u_1) \, .
$$
Then
$$
(a_1,a_2,\ldots,a_n)\succ(b_1,b_2,\ldots,b_n)
$$
so that
$$
f(a_1)+f(a_2)+ \ldots +f(a_n) \ge f(b_1)+f(b_1)+ \ldots +f(b_n)
$$
which is the desired conclusion.
Best Answer
Set $x_{n+1} = 0 $.
Hint: Prove that for any series with $x_{n+1} = 0$,
$$\sum_{i=1}^n \left[ (x_i - x_{i+1}) (\sum_{j=1}^i y_j) \right] = \sum_{j=1}^n x_j y_j. $$
Corollary: $ \sum x_i y_i - x_i^2 = \sum (x_i - x_{i+1} ) ( T_ i - S_i ) \geq 0 $, hence $ \sum x_i y_i \geq \sum x_i^2$.
The identity in the hint is known as "summation by parts", and is analogous to the "integration by parts" identity.