Given #{r,s,u,v} in RR^4# Prove that #min {rs^2,su^2,uv^2,vr^2} le 1/4#?
See explanation...
Let:
Then:
#{ (rs^21/4 = a+1/2  (b+1/2)^2  1/4 = abb^2), (s  u^21/4 = b+1/2  (c+1/2)^2  1/4 = bcc^2), (u  v^21/4 = c+1/2  (d+1/2)^2  1/4 = cdd^2), (v  r^21/4 = d+1/2  (a+1/2)^2  1/4 = daa^2) :}#
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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:

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

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}).

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*} ]

Summing all these inequalities gives: [ r  s^2 + s  u^2 + u  v^2 + v  r^2 > 1 ]

Simplifying, we get: [ r  r^2  s^2 + s  u^2 + u  v^2 + v > 1 ]

Rearranging terms, we obtain: [ (r + s + u + v)  (r^2 + s^2 + u^2 + v^2) > 1 ]

Now, we can use the CauchySchwarz Inequality to simplify: [ (r + s + u + v)^2 \leq 4(r^2 + s^2 + u^2 + v^2) ]

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 ]

Simplifying further: [ 3(r^2 + s^2 + u^2 + v^2) > 1 ]

Since the sum of squares of real numbers is always nonnegative, the lefthand side of this inequality is nonnegative.

But the righthand side is positive, which leads to a contradiction.

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

Therefore, (\min{r  s^2, s  u^2, u  v^2, v  r^2} \leq \frac{1}{4}).
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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 (CauchySchwarz 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.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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