# Prove that $\frac {x_1^3}{y_1}+\frac {x_2^3}{y_2}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)$

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 $$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_2}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)\,.$$

My idea:

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

$$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_2}+...+\frac {x_n^3}{y_n}= (x_1y_1 \frac {x_1^2}{y_1^2}+...+x_ny_n\frac {x_n^2}{y_n^2})$$

$$x_1y_1+...+x_ny_n\leq x_1^2+...+x_n^2$$ 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 ? Commented Sep 15, 2018 at 19:22
• @Ahmad Bazzi It's from hypothesis.
– rafa
Commented Sep 15, 2018 at 19:38

It is reasonable to guess the equality condition happens at $$n = 2k$$, with $$(x_i, y_i ) = (a, b)$$ for $$k$$ values, and $$(x_i, y_i) = (b, a)$$ for $$k$$ other values.
OP also observed that $$\frac{ a}{b} \leq \frac{ x_i } { y_i} \leq \frac{ b}{a}$$, and that the equality for the problem happens at these extremes. This motivates us to consider:

$$( \frac{ x_i}{y_i} - \frac{a}{b} ) ( \frac{ x_i}{y_i } - \frac{b}{a} ) \leq 0.$$

Expanding and multiplying by $$x_i y_i$$, we get

$$\frac{ x_i^3 } { y_i } + x_i y_i \leq \frac{ a^2+b^2}{ab} x_i^2.$$

Taking the sum over the indices, it looks similar to the desired inequality (so hopefully we're on the right track):

$$\sum \left( \frac{ x_i^3}{y_i} + x_i y_i \right) \leq \frac{ a^2 + b^2 } { ab} \sum x_i^2.$$

How do we deal with the additional $$\sum x_i y_i$$? We need to bound it from below (thus OP's observation that $$\sum x_i y_i \leq \sum x_i^2$$ isn't that helpful). Luckily we have Polya-Szego inequality, which states that

$$\frac{1}{4} ( \frac{ a}{b} + \frac{b}{a} ) ^2 ( \sum x_i y_i) ^2 \geq ( \sum x_i ^2 ) ( \sum y_i^2) \\ \Rightarrow (\sum x_iy_i) \geq \frac{2ab}{a^2+b^2} \sum x_i^2.$$

Hence, we have the desired inequality:

$$\sum \frac{ x_i^3 } {y_i} \leq ( \sum x_i^2 ) \left( \frac{ a^2 +b^2}{ab} - \frac{2ab}{a^2+b^2} \right) = ( \sum x_i ^2) \frac{ a^4 + b^4 } { ab(a^2+b^2 ) } .$$

Equality happens when

• $$(x_i, y_i ) = (a, b), (b, a)$$
• $$\sum x_i ^2 = \sum y_i^2$$ implies that we must have equal occurrence of each type
Hence, the equality case that we described at the start is a complete characterization.

Notes

• (See River's post) Polya's inequality (in this form) can be proven in a similar manner, $$\sum y_i^2( \frac{ x_i}{y_i} - \frac{a}{b} ) ( \frac{ x_i}{y_i } - \frac{b}{a} ) \leq 0 \Rightarrow \sum x_i^2 + y_i^2 \leq ( \frac{a}{b} + \frac{b}{a} )x_i y_i$$. Can you generalize this to the given form?
• It is very nice. A typo in the second equation. $( \frac{ x_i}{y_i} - \frac{a}{b} ) ( \frac{ x_i}{y_i } - \frac{b}{a} )$ is helpful here. And then we want a lower bound of $\sum x_iy_i$ and we will try to search the dictionary of inequalities to see if one gives lower bound of this form. Commented Sep 5 at 1:23
• By the way, did you see Andreas's answer for the case $n=3$ in the author's another question? Commented Sep 5 at 1:28
• @RiverLi Yup, I saw the linked question, alongside your solution which I didn't bother to read lol. Andreas' answer suffers from the same issue. $\quad$ I'm debating closing that question for this (but am not certain if that's the right procedure), esp since we now have a clean proof. Commented Sep 5 at 1:39
• Nice and we should wait for Andreas' reply. I think you should answer there as well (perhaps no need to use Polya-Szego inequality). I think we should not close that question since this question is a generalization of that. Commented Sep 5 at 1:42
• And yes, my solution is very complicated :) Commented Sep 5 at 1:43

Alternative proof inspired by Calvin Lin's very nice solution.

Since $$\frac{a}{b} \le \frac{x_1}{y_1} \le \frac{b}{a}$$, we have $$\left(\frac{x_1}{y_1} - \frac{a}{b}\right)\left(\frac{x_1}{y_1} - \frac{b}{a}\right)\left(x_1y_1 + \frac{ab}{a^2 + b^2}y_1^2\right) \le 0$$ which is written as $$\frac{x_1^3}{y_1} \le \frac{a^2 + b^2}{ab} x_1^2 - \frac{ab}{a^2 + b^2}(x_1^2 + y_1^2).$$ Thus, we have \begin{align*} \sum_i \frac{x_i^3}{y_i} &\le \frac{a^2 + b^2}{ab} \sum_i x_i^2 - \frac{ab}{a^2 + b^2}(\sum_i x_i^2 + \sum_i y_i^2)\\ &= \frac{a^4+b^4}{ab (a^2+b^2)}\sum_i x_i^2. \end{align*}

We are done.

• Haha, yea that's combining with the proof of Polya. I decided this presentation because it distracts away from the key idea of the first inequality, which has been commonly used in inequalities of this form. Commented Sep 6 at 21:23
• @CalvinLin When I was looking through your idea, I found that the form of $xu \ge \cdots$ can also be derived from $\left(\frac{x_1}{y_1} - \frac{a}{b}\right)\le 0$. Combining them, we get the bound. My idea is not to prove the Polya Szego (perhaps it actually is). I first dealt with the $n=3$ case. We need to prove that $xu + yv + zw \ge \frac{ab}{a^2 + b^2}(x^2 + y^2 + z^2 + u^2 + v^2 + w^2)$. Then I found this inequality is separable (or localization, or sth. like isolated fudging), that is, it suffices to prove that $xu \ge \frac{ab}{a^2 + b^2}(x^2 + u^2)$ for all $x, u \in [a, b]$. Commented Sep 7 at 0:30
• @CalvinLin (Continued) The Polya Szego seems to consider the $xu + yv + zw$ as whole, rather than localization? By the way, as an afterthought, the use of $\left(\frac{x_1}{y_1} - \frac{a}{b}\right)\left(\frac{x_1}{y_1} - \frac{b}{a}\right)\le 0$ is commonly used in inequality proofs, but I wasn't able to think of the use of it for this problem. Commented Sep 7 at 0:35

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.

• While I agree that "By the rearrangement inequality, the maximum value of the LHS is obtained if the highest factors $q_i$ are chosen for the highest $x_i^2$", I disagree with "This can be achieved by letting as many $y_i = a$ for the highest indices", as you're not allowed to change the value of the variables, only permute them. You need a separate (smoothing?) argument to explain why you can set $y_n = a$. EG If $\sum y_i ^2 > (n-1) b^2 + a^2$, then we'd be unable to set $y_n = a$. So this solution, to me, has a huge hole in it. Commented Sep 4 at 22:44