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) $

Question

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 $\frac {a^4+b^4}{ab (a^2+b^2)}$.

Answer 1

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?

Answer 2

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.

Answer 3

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.