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? 
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.

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 
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315  
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 CauchySchwartz or AMGM 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?

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 $\displaystyle t=1$ . 

Tags 
calculus, minimization, problem, sort 
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