Prove that inequality Hardy inequality

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

• 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. – Andreas Jun 19 at 12:21

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.

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