February 5th, 2019, 11:38 AM  #1 
Senior Member Joined: Sep 2015 From: USA Posts: 2,314 Thanks: 1230  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, 03:17 PM  #2 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
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 03:39 PM. 
February 5th, 2019, 06:12 PM  #3 
Senior Member Joined: Feb 2010 Posts: 702 Thanks: 137 
Google something called the Purkiss Principle.

February 7th, 2019, 11:33 AM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
Then how to prove the equality ?

February 7th, 2019, 12:47 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,314 Thanks: 1230  
February 7th, 2019, 05:49 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,298 Thanks: 1971 
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, 05:54 PM  #7 
Senior Member Joined: Sep 2016 From: USA Posts: 559 Thanks: 324 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, 04:39 AM  #8 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
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 04:45 AM. 
February 8th, 2019, 05:28 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,298 Thanks: 1971 
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, 08:14 AM  #10 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
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 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Question: Percentage word problem..sort of  extremistpullup  Elementary Math  6  July 9th, 2011 10:56 PM 
Minimization problem 2  Fantini  Calculus  2  October 2nd, 2010 09:04 AM 
Minimization problem  Fantini  Calculus  4  September 30th, 2010 05:42 PM 
Geometric Minimization problem  @nthony  Algebra  4  May 9th, 2009 04:24 PM 
Minimization problem  nsa_tanay  Applied Math  1  November 24th, 2008 12:25 PM 