Origin of coefficients in the inequality $(a- c)^{2}+ (b- d)^{2}\geq\frac{7}{9}ab- \frac{7}{20}(c^{2}+ 4d^{2})$

Question

Given four real numbers $a, b, c, d$, Ji Chen proved the following inequality (see AoPS source):

$$(a - c)^2 + (b - d)^2 \geq \frac{7}{9} ab - \frac{7}{20}(c^2 + 4d^2)$$

The original problem assumed the condition:

$$ab = c^2 + 4d^2 = 4$$

and hid the right-hand side of the inequality. I wondered how Ji Chen could have discovered those constants $\frac{7}{9}$ and $-\frac{7}{20}$. From what I understand, he may have used a discriminant method to derive this result.

While searching the AoPS forums for insights, I came across another impressive result by user @ye109, who generalized the inequality to:

$$(a - c)^2 + (b - d)^2 \geq \frac{2(k^4 - 4)}{3k^2} ab + \frac{k^4 - 4}{k^4 - 16}(c^2 + 4d^2)$$

where the best possible value of $k \approx 1.672955253$ is the real root of the polynomial:

$$k^{12} - 28k^8 - 36k^6 - 128k^4 + 1024 = 0$$

This result seems to offer a parametric generalization. I'm curious: If Ji Chen used a discriminant approach to derive his constants, is it possible that this more general inequality was also derived using a discriminant — perhaps extended to a polynomial setting, not just numerical constants?

My question:

• How can one systematically use the discriminant method (or a similar technique) to derive such constants in inequalities?
• Could the coefficients in @ye109’s result also have arisen from a polynomial discriminant condition?
• What’s a good way of thinking to transition from Ji Chen’s specific inequality to this parametric generalization?

I’d really appreciate any insights or suggested strategies for tackling problems like this. Thank you so much!

Answer 1

it's about a symmetric matrix, the Hessian of your quadratic form, being positive semidefinite.

$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 7 }{ 18 } & 1 & 0 & 0 \\ \frac{ 324 }{ 275 } & \frac{ 126 }{ 275 } & 1 & 0 \\ \frac{ 18 }{ 5 } & \frac{ 12 }{ 5 } & \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 180 & - 70 & - 180 & 0 \\ - 70 & 180 & 0 & - 180 \\ - 180 & 0 & 243 & 0 \\ 0 & - 180 & 0 & 432 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 7 }{ 18 } & \frac{ 324 }{ 275 } & \frac{ 18 }{ 5 } \\ 0 & 1 & \frac{ 126 }{ 275 } & \frac{ 12 }{ 5 } \\ 0 & 0 & 1 & \frac{ 8 }{ 3 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 180 & 0 & 0 & 0 \\ 0 & \frac{ 1375 }{ 9 } & 0 & 0 \\ 0 & 0 & \frac{ 1701 }{ 55 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 7 }{ 18 } & 1 & 0 & 0 \\ - 1 & - \frac{ 126 }{ 275 } & 1 & 0 \\ 0 & - \frac{ 324 }{ 275 } & - \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 180 & 0 & 0 & 0 \\ 0 & \frac{ 1375 }{ 9 } & 0 & 0 \\ 0 & 0 & \frac{ 1701 }{ 55 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 7 }{ 18 } & - 1 & 0 \\ 0 & 1 & - \frac{ 126 }{ 275 } & - \frac{ 324 }{ 275 } \\ 0 & 0 & 1 & - \frac{ 8 }{ 3 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 180 & - 70 & - 180 & 0 \\ - 70 & 180 & 0 & - 180 \\ - 180 & 0 & 243 & 0 \\ 0 & - 180 & 0 & 432 \\ \end{array} \right) $$

See https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia

Answer 2

$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq Uab-V \left ( c^{2}+ 4d^{2} \right )$$ (for all reals $a,b,c,d$)

precisely when $$ V \geq 0$$ and $$ |U| \leq \frac{4V}{\sqrt{(1+V)(1+4V)}} $$

In turn, the largest $U$ is rational when $$ V = \frac{m^2 - n^2}{4n^2 - m^2} \geq 0$$ where we get $$ |U| \leq 4\frac{|m^2-n^2|}{3mn} $$

In particular, when we take $n=3,m=4$ we get $V= 7/20$ and the largest $|U|$ that still gives positive semidefinite is $7/9$

The positive semidefinite matrix with extremal $|U|$

using $2n > m > n,$ $$ \left( \begin{array}{cccc} 1&\frac{2m^2-2n^2}{3mn}&-1&0\\ \frac{2m^2-2n^2}{3mn}&1&0&-1\\ -1&0&\frac{3n^2}{4n^2-m^2}&0\\ 0&-1&0&\frac{3m^2}{4n^2-m^2}\\ \end{array} \right) $$