Suppose $n$ is a positive integer, $2n$ reals $x_i, y_i (1\le i \le n) $ satisfy $$\sum_{i=1}^n x_i^2 = \sum_{i=1}^n y_i^2 = 1.$$ positive reals $0 < \lambda_1 \le \lambda_2 \le ... \le \lambda_n, \ 1 \in [\lambda_1, \lambda_n].$ Prove that $$\lambda_1 \le \sum_{i=1}^n \lambda_ix_i^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 - \frac{(\sum_{i=1}^n\lambda_ix_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \le \lambda_n.$$

It seems stronger than Cauchy-Schwarz inequality. $$\sum_{i=1}^{n}\lambda_{i}x^2_{i}\sum_{i=1}^{n}\lambda_{i}y^2_{i}\ge (\sum_{i=1}^{n}\lambda_{i}x_{i}y_{i})^2$$

Idear 2: I try to use From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$, $$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$

But I can't it.Thanks

  • $\begingroup$ I wonder about the "Hardy inequality" in the title as I cannot see the relation to Hardy's inequality. Can you explain? I would rather say something in the title about the relation to (weighted) Cauchy-Schwarz inequality, to make search algorithms find this nice question. $\endgroup$
    – Andreas
    Jun 19, 2019 at 12:21

1 Answer 1


EDIT (2019-06-19): Here is a complete result.

Recall from the classical proof of the Cauchy-Schwarz-inequality, that there exists the equality $$ \sum_{i=1}^n \lambda_ix_i^2 - \frac{(\sum_{i=1}^n\lambda_ix_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} = \sum_{i=1}^n \lambda_i \left[ x_i - a y_i \right]^2 $$ with $$ a = \frac{\sum_{i=1}^n\lambda_ix_iy_i}{\sum_{i=1}^n\lambda_iy_i^2} $$

Hence the claim can be rewritten (introducing $F(x_i,y_i,\lambda_i)$) as: $$ \lambda_1 \le F(x_i,y_i,\lambda_i) =\sum_{i=1}^n \lambda_i \left[ x_i - a y_i \right]^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 \le \lambda_n. $$

For the left inequality, we note that $0 \le \lambda_1 \le \lambda_i$ and $0 \le \lambda_1 \le 1$, hence it suffices to show: \begin{align} \lambda_1& \le \lambda_1 \sum_{i=1}^n \left[ x_i - a y_i \right]^2 + \lambda_1 \left(\sum_{i=1}^n x_iy_i\right)^2 \\ \Longleftrightarrow 1 &\le \sum_{i=1}^n \left[ x_i - a y_i \right]^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 \\ &= \sum_{i=1}^n \left[ x_i^2 - 2 a x_iy_i + a^2y_i^2\right] + \left(\sum_{i=1}^n x_iy_i\right)^2 \\ &=1 + a^2 -2a \sum_{i=1}^n x_iy_i +\left(\sum_{i=1}^n x_iy_i\right)^2 \\ &=1 + (a - \sum_{i=1}^n x_iy_i )^2 \end{align} and this establishes the left inequality.

For the right inequality we do the following. Let $\sum_{i=1}^n x_iy_i = q$. Replace $x_i$ with $x_i = q y_i + n_i$. The reason to call the new variable $n_i$ is that $\sum_{i=1}^n y_i n_i = \sum_{i=1}^n y_i (x_i - q y_i) = q - q \sum_{i=1}^n y_i^2 = 0$, so the $(n_i)$ can be understood as the vector component of the vector $x$ which is normal (hence the n) to the $y$-vector. We have $\sum_{i=1}^n n_i^2 = \sum_{i=1}^n (x_i - q y_i)^2 = 1 - 2q^2 +q^2 = 1 - q^2$, which will be used below.

With this replacement, the expression in question becomes \begin{align} F(x_i,y_i,\lambda_i) &= \sum_{i=1}^n \lambda_i(q y_i + n_i)^2 + q^2 - \frac{(\sum_{i=1}^n\lambda_i(q y_i + n_i)y_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= q^2\sum_{i=1}^n \lambda_i y_i ^2 + 2q \sum_{i=1}^n \lambda_i y_i n_i + \sum_{i=1}^n \lambda_i n_i^2 + \\ &\qquad + q^2- \frac{(q \sum_{i=1}^n\lambda_i y_i^2 + \sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= q^2\sum_{i=1}^n \lambda_i y_i ^2 + 2q \sum_{i=1}^n \lambda_i y_i n_i + \sum_{i=1}^n \lambda_i n_i^2 + \\ &\qquad + q^2- q^2 \sum_{i=1}^n\lambda_i y_i^2 - 2q \sum_{i=1}^n\lambda_in_iy_i -\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= \sum_{i=1}^n \lambda_i n_i^2 + q^2 -\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \end{align} Structurally, this looks strikingly similar to the original formulation. The difference (which we will exploit) is that the vector $(n_i)$ has a relation to the $q$, which was not present before.

Replacing $q^2 = 1 - \sum_{i=1}^n n_i^2 $ (we had computed that already above) and bounding $\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \ge 0$ gives \begin{align} F(x_i,y_i,\lambda_i) &\le \sum_{i=1}^n \lambda_i n_i^2 + 1 - \sum_{i=1}^n n_i^2 \\ &\le \lambda_n \sum_{i=1}^n n_i^2 + 1 - \sum_{i=1}^n n_i^2 \\ &= 1 + (\lambda_n - 1)\sum_{i=1}^n n_i^2 \end{align} Further, we have that $\lambda_n - 1 \ge 0 $ and $\sum_{i=1}^n n_i^2 = 1 -q^2 \le 1$, so we can conclude $$ F(x_i,y_i,\lambda_i) \le 1 + (\lambda_n - 1) = \lambda_n $$ which is the desired result for the right inequality. This completes the proof. $\qquad \square$

Some interpretation: The bounding $\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \ge 0$ gives rise to the conclusion that this term is "not important" so it could be bounded away. This is the case, since without the $\lambda_i$, this term would become zero, as $\sum_{i=1}^n n_iy_i = 0$. Indeed, few computer simulations show that, for various choices of the $\lambda_i$, the maximum of the expression in question will be obtained when the vector $x$ is chosen almost perfectly perpendicular to the vector $y$, hence $x \simeq n$, which means that the discussed bounding is "save" as the term becomes small and thus doesn't produce much difference to the true result.

  • $\begingroup$ In the first line of the last chain of expressions, how did you replace $\lambda_i$ with $1$ in the first summation? $\endgroup$
    – Explorer
    Jun 18, 2019 at 10:51
  • $\begingroup$ @GeethuJoseph I put that explanation in the main text $\endgroup$
    – Andreas
    Jun 18, 2019 at 14:30
  • $\begingroup$ Nice proof! :) +1 $\endgroup$
    – Explorer
    Jun 19, 2019 at 6:50

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