So this is what I am trying to prove $\frac{(\sum_i w_i x_i)}{(\sum_i w_i y_i^2)} \leq \sum_i w_i \frac{x_i}{y_i^2}$, where $w_i \geq 1$, $x_i>0$, and $y_i>0$.
I have no idea where to begin with, I have scoured many inequalities like Cauchy–Schwarz inequality, Titu's lemma or to begin like this $\frac{\sum_i x_i}{\sum_i y_i} \leq \max_i\frac{x_i}{y_i}$. I am not even sure this lemma can be prove. So any help is appreciated.
Best Answer
Since $w_i\geqslant 1$ for all $i$, for non-negative numbers $t_i$, \begin{align*}\Bigl(\sum_{i=1}^\infty w_it_i\Bigr)^2 & =\sum_{i,j=1}^\infty w_iw_jt_it_j = \sum_{i=1}^\infty w_i^2t_i^2 +\sum_{i=1}^\infty\sum_{i\neq j=1}^\infty w_iw_jt_it_j \\ & \geqslant \sum_{i=1}^\infty w_i^2t_i^2 \geqslant \sum_{i=1}^\infty w_it_i^2. \end{align*}
Taking square roots gives $$\Bigl(\sum_{i=1}^\infty w_it_i^2\Bigr)^{1/2}\leqslant \sum_{i=1}^\infty w_it_i.$$ This is just a special case of the fact that the weighted $\ell_p(w)$ norm is not less than the weighted $\ell_q(w)$ norm when $1\leqslant p<q$ when all weights are at least $1$.
EDIT: This paragraph uses Cauchy-Schwarz on the weighted Hilber space $\ell_2(w)$ whose inner product is $\langle (a_i),(b_i)\rangle=\sum_i a_ib_i$. In this case, $\|a\|2=\langle a,a\rangle=\sum_i w_i|a_i|^2$. The Cauchy-Schwarz inequality is valid for any inner product, including the weighted one, and says $|\langle a,b\rangle|\leqslant \|a\|\|b\|$).
Using Cauchy-Schwarz and then this inequality, we have \begin{align*}\sum_{i=1}^\infty w_ix_i & = \sum_{i=1}^\infty w_i y_i^2 \cdot (x_i/y_i^2) \leqslant \Bigl(\sum_{i=1}^\infty w_i (y_i^2)^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^\infty w_i(x_i/y_i^2)^2\Bigr)^{1/2} \\ & \leqslant \Bigl(\sum_{i=1}^\infty w_iy_i^2\Bigr)\Bigl(\sum_{i=1}^\infty w_i(x_i/y_i^2)\Bigr).\end{align*}
Below I'll give a proof that usual Cauchy-Schwarz on the usual inner product, rather than the weighted one. But it's a good idea to know about the full strength and realm of applicability of Cauchy-Schwarz.
Using Cauchy-Schwarz and then this inequality, we have \begin{align*}\sum_{i=1}^\infty w_ix_i & = \sum_{i=1}^\infty \sqrt{w_i} y_i^2 \cdot (\sqrt{w_i}x_i/y_i^2) \leqslant \Bigl(\sum_{i=1}^\infty (\sqrt{w_i} y_i^2)^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^\infty (\sqrt{w_i}x_i/y_i^2)^2\Bigr)^{1/2} \\ & = \Bigl(\sum_{i=1}^\infty w_i(y_i^2)^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^\infty w_i(x_i/y_i^2)^2\Bigr)^{1/2}\leqslant \Bigl(\sum_{i=1}^\infty w_iy_i^2\Bigr)\Bigl(\sum_{i=1}^\infty w_i(x_i/y_i^2)\Bigr).\end{align*}