My Math Forum Not exactly calculus but it is a sort of minimization problem
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 February 5th, 2019, 10:38 AM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 Not exactly calculus but it is a sort of minimization problem $\text{given }a,b,c >0 \text{ show that}$ $\dfrac{a^3}{b^2(5a+2b)}+\dfrac{b^3}{c^2(5b+2c)}+ \dfrac{c^3}{a^2(5c+2a)} \geq \dfrac 3 7$ The equality occurs when $a=b=c$ but how to show this is a minima? Thanks from topsquark and idontknow
 February 5th, 2019, 02:17 PM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 By simple math , it is seen that the fractions have the same minimal value . One of them is $\displaystyle F(a,b)=\frac{a^3 }{b^2 (5a+2b)}$ . (by calculus find the values ) To make it easier substitute for $\displaystyle x=\frac{a}{b}\; \;$ then $\displaystyle F(a,b)=\frac{x^3 }{5x+2 } \; \; , x>0$ But I think this is not a method , maybe someone else can assist. Last edited by idontknow; February 5th, 2019 at 02:39 PM.
 February 5th, 2019, 05:12 PM #3 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 140 Google something called the Purkiss Principle. Thanks from idontknow
 February 7th, 2019, 10:33 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 Then how to prove the equality ?
February 7th, 2019, 11:47 AM   #5
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Quote:
 Originally Posted by idontknow Then how to prove the equality ?
equality is trivial, set a=b=c

 February 7th, 2019, 04:49 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,633 Thanks: 2080 If $x = a/b$ and $y = b/c$, the expression is a function of $x$ and $y$ for which the minimum (for positive $x$ and $y$) can be found to occur when $x = y = 1$ with the help of calculus (partial differentiation). This should work, but might be a bit tedious to do.
 February 7th, 2019, 04:54 PM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics One way to rigorously show that it is a minimizer is to check whether the Hessian matrix is positive definite. While this is guaranteed to succeed, its not very enlightening and is almost certainly not the "best" solution. Problems with this sort of symmetry are almost always some clever trick using Cauchy-Schwartz or AM-GM inequality. The change of variables proposed by Idontknow appled to all 3 fractions is probably a step in the right direction.
 February 8th, 2019, 03:39 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 No need to know the values of a,b,c or x,y,z . $\displaystyle F=\frac{x^3}{5x+2} + \frac{y^3 }{5y+2 } +\frac{z^3}{5z+2 }$ . And 3/7 can be the minimal , if not then it is lower then minimal , so it means finding the minimal value proves the equality . Since a constraint is given , it takes time to use calculus . So the minimal occurs at the same value for the three fractions , x=y=z . Now there are many ways to get $\displaystyle a=b=c$ from $\displaystyle a/b=b/c=c/a \;$ for a,b,c > 0 . Last edited by idontknow; February 8th, 2019 at 03:45 AM.
 February 8th, 2019, 04:28 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,633 Thanks: 2080 If you don't obtain any specific values or use calculus, how would you know whether you've found a minimum, as distinct from a maximum? Thanks from idontknow
 February 8th, 2019, 07:14 AM #10 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 The minimum occurs in a point $\displaystyle t=x=y=z$ . a=bt ; b=ct ; c=at ; and by the sum $\displaystyle a+b+c =t(a+b+c) \; \;$ or $\displaystyle t=1$ .

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