March 2nd, 2019, 11:05 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88  Inequality proof
$\displaystyle a^2 +b^2 +c^2 +d^2 =1 \; \; $ , where $\displaystyle a,b,c,d >0$ . Prove that $\displaystyle a+b+c+d 1 \geq 16abcd$ . 
March 4th, 2019, 02:10 PM  #2 
Member Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 
I don't know if this helps but the sum of the squares of the sides of any parallelogram is equal to the sum of the squares of the diagonals. In your case, a^2 + b^2 + c^2 + d^2 is the sum of the squares of the sides of any parallelogram and the sum of the squares of the diagonals is equal to 1. The condition a,b,c, and d must satisfy makes "any" parallelogram not true but there is one that meets the criteria for a,b positive Let a=c and b=d, then a^2 + b^2 = 1/2 All my results are a bit meaningless and pointless really. Nothing worth showing. 
March 4th, 2019, 02:39 PM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
I almost proved it in a weak way : $\displaystyle 16abcd\leq 1 =a^2 +b^2 +c^2 +d^2 \leq a+b+c+d $ . But the inequality contains a 1 to the right side . 
March 4th, 2019, 03:02 PM  #4 
Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry 
I've been working at this for a while now and that (1) has messed me up every time. I do know that $a+b+c+d \in (1,2]$ and $16abcd \in (0,1]$, but I can't seem to link them together to account for the (1). Since the constraint is a sphere maybe there would be a way to do some weird trigonometry on it?
Last edited by skipjack; March 4th, 2019 at 04:06 PM. 
March 4th, 2019, 03:04 PM  #5 
Member Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3  
March 4th, 2019, 03:25 PM  #6 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
Yes the constraint is a sphere and trigonometry may be useful . Another way is to use Lagrangemultipliers . 

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