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March 2nd, 2019, 11:05 AM   #1
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Inequality proof

$\displaystyle a^2 +b^2 +c^2 +d^2 =1 \; \; $ , where $\displaystyle a,b,c,d >0$ .
Prove that $\displaystyle a+b+c+d -1 \geq 16abcd$ .
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March 4th, 2019, 02:10 PM   #2
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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.
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March 4th, 2019, 02:39 PM   #3
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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 .
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March 4th, 2019, 03:02 PM   #4
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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?
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Last edited by skipjack; March 4th, 2019 at 04:06 PM.
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March 4th, 2019, 03:04 PM   #5
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Quote:
Originally Posted by idontknow View Post
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.
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March 4th, 2019, 03:25 PM   #6
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Yes the constraint is a sphere and trigonometry may be useful .
Another way is to use Lagrange-multipliers .
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