Prove that if ( are positive reals) then
.
Probably. (I haven't done it so don't know but it's from a series...) Maybe the inequality could be used as we proved it.
This problem has been gripping me for a while now and still I can’t solve it. Here’s what I’ve done with the inequality so far. Multiply through by a+b+c:
Now
Thus
Let . Then and the inequality becomes
This is what needs to be proved. I think the trick is to let , where , but I still can’t do it.
This problem has also got my head interacting with brick walls.
Following JaneBennet's ideas, we want to show that if A, B, C are positive numbers with BC+CA+AB=1, then
.
Let be the cubic equation with roots A, B and C. Then p=A+B+C, q=BC+CA+AB and r=ABC. Hence q=1.
The equation has roots Ap, Bp and Cp. In other words, the equation has roots A(A+B+C), B(A+B+C), C(A+B+C).
Therefore the equation has roots A(A+B+C)+2, B(A+B+C)+2, C(A+B+C)+2. In other words (multiplying out the brackets), the equation has those roots.
Replacing x by 1/x, we see that the equation has roots , , .
The sum of the roots of this cubic equation is also the coefficient of x^2 divided by the coefficient of x^3. So to prove our inequality we need to show that , where p=A+B+C and r=ABC. An equivalent inequality is .
Now what?
Okay, I can complete the proof of this, but it's not at all illuminating. There must be a more conceptual explanation for this strange inequality.
The situation is that A, B, C are positive numbers with BC+CA+AB=1. If p=A+B+C and r=ABC then we want to show that . Since the proof involves calculus, I'll change the notation (again!) and write x, y, z for A, B, C.
So x, y, z are positive reals with yz+zx+xy=1, and we want to show that . Let . Then we want to show that the minimum value of f on D is 4.
The boundary of D consists of points for which one of the coordinates is zero. Then the other two coordinates must be w and 1/w, for some positive number w. The value of f at this point is , and it is easy to see that the minimum of this expression is 4 (occurring when w=1).
To find the extrema of f in the interior of D, use the method of Lagrange multipliers, and find the zeros of the partial derivatives of g(x,y,z)=f(x,y,z)+λ(yz+zx+xy-1):
,
,
.
Subtract the last of these from the middle one, to get (and two similar equations with x,y and z cyclically interchanged). If x,y and z are all different then , which is impossible (since it implies that x, y and z are all equal).
Therefore the extreme values of f on D must occur at points where at least two of x, y and z are equal. Suppose for example that z=y. Then the conditon (x,y,z)∈D becomes y(2x+y)=1, and so . Note that this implies (otherwise x would be negative). Also, and .
Then , which simplifies(?) to . We want this to be greater than or equal to 4, or in other words . When the term 64y^2 is taken over to the left side, this factorises as , and this is clearly true when 0<y≤1.
By the way, some people think this problem is easy.
Art of Problem Solving Forum
Unfortunately I don’t understand the steps in their solution.