Question about "Reduce"
- To: mathgroup at smc.vnet.net
- Subject: [mg61888] Question about "Reduce"
- From: "Stanley Rabinowitz" <stan.rabinowitz at comcast.net>
- Date: Thu, 3 Nov 2005 04:59:02 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Consider the inequality:
ineq = L(1+u^2)(1+v^2)<=u+v+M u v(u v-1)
I am trying to find all real values of L and M that make this
inequality
true for all real u and v subject to the constraints:
0 <= u <= Sqrt[3]
0 <= v <= Sqrt[3]
0 <= (1 - u v)/(u + v) <= Sqrt[3]
I thought I could do this with Reduce.
I issue the command
Reduce[ForAll[{u, v}, 0 <= u <= Sqrt[3] && 0 <= v <= Sqrt[3] && 0 <= (1
- u v)/(u + v)<= Sqrt[3], ineq],
{L,M}, Reals] // FullSimplify
and I get back a result equivalent to
L<=0 && M <= 3Sqrt[3]-8L
Am I doing something wrong?
It is telling me that L must be less than 0.
But I know for a fact that
M = 1 + Sqrt[3] and L = (2Sqrt[3] - 1)/8 work
and in this case, L>0.
I can verify that these value of L and M work by issuing the command
Reduce[ForAll[{u, v}, 0 <= u <= Sqrt[3] && 0 <= v <= Sqrt[3] && 0 <= (1
- u v)/(u + v)<= Sqrt[3],
ineq/. {M -> 1 + Sqrt[3], L -> (2Sqrt[3]-1)/8}]]
which returns the value "True".
So why is it telling me L<=0?