# $\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)$

Let $b>a>0$ and $x_1, x_2,\ldots,x_n,y_1, y_2,\ldots,y_n\in [a,b]$. If $$x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2\,,$$ then $$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)\,.$$

My idea:

I proved that $\dfrac {a}{b}\leq \frac {x_k}{y_k}\leq \dfrac {b}{a}$.

$$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+...+\frac {x_n^3}{y_n}= (x_1y_1 \frac {x_1^2}{y_1^2}+...+x_ny_n\frac {x_n^2}{y_n^2})$$

$x_1y_1+...+x_ny_n\leq x_1^2+...+x_n^2$ by Cauchy Schwartz.

Unfortunately $\frac {x_k}{y_k}$ is not smaller than $\frac {a^4+b^4}{ab (a^2+b^2)}$.

• is $x_1^2+x_2^2+....+x_n^2=y_1^2+y^2+...+y_n^2$ one of the assumptions, or you just assumed it ? – Ahmad Bazzi Sep 15 '18 at 19:22
• @Ahmad Bazzi It's from hypothesis. – rafa Sep 15 '18 at 19:38

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.