Tighter version of Cauchy-Schwarz?

I checked numerically that $$\left( \sum_N p_N \dfrac{C_N}{D_N^2} \right) \left( \sum_N p_N D_N \right) \geq \left( \sum_N p_N \dfrac{C_N}{D_N} \right)$$ where $$\sum_N p_N = 1 \;, \quad 0\leq p_N\leq 1 \;, \quad 0\leq \dfrac{C_N}{D_N} \leq 1 \;,$$ but I'm not able to prove it analytically. Using Cauchy-Schwarz gives the bound $$\left( \sum_N p_N \dfrac{C_N}{D_N^2} \right) \left( \sum_N p_N D_N \right) \geq \left(\sum_N \sqrt{p_N \dfrac{C_N}{D_N^2} p_N D_N}\right)^2 = \left(\sum_N p_N \sqrt{\dfrac{C_N}{D_N}} \right)^2$$ which is smaller than $$\left( \sum_N p_N \dfrac{C_N}{D_N} \right)$$.

Is there a way to prove my first inequality?

• @rhombicosicodecahedron your inequality is wong.
– m137
May 1, 2019 at 6:54

Smallest counter-example:

• $$p_1 = p_2 = 1/2$$

• $$C_1 = 0, C_2 = 1$$

• $$D_1, D_2 > 0$$ and exact values to be determined later

• $$LHS = {1 \over 2 D_2^2} {D_1 + D_2 \over 2} = {1 \over 4} {D_1 + D_2 \over D_2^2}$$

• $$RHS = {1 \over 2 D_2} = {1 \over 4}{D_2 + D_2 \over D_2^2},$$ so RHS > or = or < LHS depending on $$D_2$$ > or = or < $$D_1$$. Taking $$D_2 > D_1$$ gives a counter-example.

More abstractly, interpreting $$p_i$$ as probabilities and $$C,D$$ as random variables, $$LHS = E[{C\over D^2}] E[D]$$ and $$RHS =E[{C\over D^2} D] = E[{C\over D}],$$ so

$$RHS - LHS =Cov({C\over D^2}, D)$$

and the counter-example is simply a case where, when $$C/D^2$$ increases, $$D$$ also increases, so the covariance is positive.

• Thanks a lot for the effort. The interpretation as probability distributions is indeed the one of my interest. I think in all cases I need to consider, $C\propto D$, and therefore the covariance is negative. In this case my conjecture can be said to be correct.
– m137
May 2, 2019 at 8:09
• Do you really mean $C \propto D$, i.e. $C/D=$ constant? Or do you simply mean as $C$ increases, $D$ also increase? In the latter case, $C$ and $D$ may be positively correlated but $C/D^2$ may not be. May 2, 2019 at 13:46

I don't think your conjecture is true. Although you might have finite sums in mind, you can test your hypothesis by summing over all $$N \in \mathbb N$$ with $$p_N = \left( \frac 12 \right)^N\quad C_N = \left( \frac 15 \right)^N\quad D_N = \left( \frac 45 \right)^N.$$

• Thank you for this example, it looks indeed falsifying my conjecture. I definitely consider finite sums, but it might also be that there ore other hidden assumptions entering in the coefficients I generate... I'll look for these!
– m137
May 2, 2019 at 7:57