Given #{r,s,u,v} in RR^4# Prove that #min {r-s^2,s-u^2,u-v^2,v-r^2} le 1/4#?

Question

Charles Anctil · Accepted Answer

See explanation...

First note that if #r = s = u = v = t# then we are seeking to maximise: #t-t^2 = 1/4-(1/4-t+t^2) = 1/4-(t-1/2)^2# so the maximum #1/4# occurs when #t = 1/2# Let: #{ (a = r-1/2), (b = s-1/2), (c = u-1/2), (d = v-1/2) :}# Then: #{ (r-s^2-1/4 = a+1/2 - (b+1/2)^2 - 1/4 = a-b-b^2), (s - u^2-1/4 = b+1/2 - (c+1/2)^2 - 1/4 = b-c-c^2), (u - v^2-1/4 = c+1/2 - (d+1/2)^2 - 1/4 = c-d-d^2), (v - r^2-1/4 = d+1/2 - (a+1/2)^2 - 1/4 = d-a-a^2) :}# Note that #a^2 >= 0#, #b^2 >= 0#, #c^2 >= 0# and #d^2 >= 0#, so if all of the above expressions are non-negative, then: #a >= b >= c >= d >= a# Hence #a=b=c=d# and we are back to #r = s = u = v#

Emilia Aguilar · Answer

undefined To prove that $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} \leq \frac{1}{4}$ for $r, s, u, v \in \mathbb{R}^4$, we can proceed as follows:

1. We'll assume the contrary, that is, suppose $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} > \frac{1}{4}$.

2. Since this minimum value is greater than $\frac{1}{4}$, it follows that each of the individual terms in the set is also greater than $\frac{1}{4}$.

3. Therefore, we have:
   $$
   \begin{align*}
   &r - s^2 > \frac{1}{4} \
   &s - u^2 > \frac{1}{4} \
   &u - v^2 > \frac{1}{4} \
   &v - r^2 > \frac{1}{4}
   \end{align*}
   $$

4. Summing all these inequalities gives:
   $$
   r - s^2 + s - u^2 + u - v^2 + v - r^2 > 1
   $$

5. Simplifying, we get:
   $$
   r - r^2 - s^2 + s - u^2 + u - v^2 + v > 1
   $$

6. Rearranging terms, we obtain:
   $$
   (r + s + u + v) - (r^2 + s^2 + u^2 + v^2) > 1
   $$

7. Now, we can use the Cauchy-Schwarz Inequality to simplify:
   $$
   (r + s + u + v)^2 \leq 4(r^2 + s^2 + u^2 + v^2)
   $$

8. Substituting this inequality into the previous expression, we get:
   $$
   4(r^2 + s^2 + u^2 + v^2) - (r^2 + s^2 + u^2 + v^2) > 1
   $$

9. Simplifying further:
   $$
   3(r^2 + s^2 + u^2 + v^2) > 1
   $$

10. Since the sum of squares of real numbers is always non-negative, the left-hand side of this inequality is non-negative.

11. But the right-hand side is positive, which leads to a contradiction.

12. Hence, our initial assumption that $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} > \frac{1}{4}$ must be false.

13. Therefore, $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} \leq \frac{1}{4}$.

Scarlett Allison · Answer

undefined To prove that $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} \leq \frac{1}{4}$ for $r, s, u, v \in \mathbb{R}^4$, we can consider the maximum value that any of the terms in the set can take.

Let's assume, for the sake of contradiction, that $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} > \frac{1}{4}$. This implies that each of the terms in the set is greater than $\frac{1}{4}$.

Without loss of generality, let's consider $r - s^2 > \frac{1}{4}$. This means $r > \frac{1}{4} + s^2$. Similarly, we can derive that $s > \frac{1}{4} + u^2$, $u > \frac{1}{4} + v^2$, and $v > \frac{1}{4} + r^2$.

Adding all these inequalities together, we get:

$r + s + u + v > \frac{1}{4} + s^2 + \frac{1}{4} + u^2 + \frac{1}{4} + v^2 + \frac{1}{4} + r^2$.

Simplifying, we have:

$r + s + u + v > \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + s^2 + u^2 + v^2 + r^2$.

$r + s + u + v > 1 + (s^2 + u^2 + v^2 + r^2)$.

However, this contradicts the fact that the sum of any four real numbers squared is always greater than or equal to 1 (Cauchy-Schwarz inequality). Therefore, our assumption that $\min\{r - s^2, s - u^2, u - v^2, v - r^2\} > \frac{1}{4}$ is incorrect, and the statement is proved.