My Math Forum Inequality proof

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 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. Thanks from idontknow
 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? Thanks from idontknow Last edited by skipjack; March 4th, 2019 at 04:06 PM.
March 4th, 2019, 03:04 PM   #5
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Quote:
 Originally Posted by idontknow 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 .
No way.

Let's say you don't know what they're asking you to show. You must derive it.

 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 Lagrange-multipliers .

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