# Prove that for any positive numbers p and q: p + q = sqrt(4pq)

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#1
I did (p + q)^2 = p^2 + 2pq + q^2 g ≥ 4pq
so p^2 -2qr + q^2 ≥ 0 which is true since it's equal to (p-q)^2 and any number squared is ≥ 0.
Since (p - q)^2 + 4pq = (p + q)^2 and (p - q)^2 ≥ 0
(p + q)^2 ≥ 4pq so
p + q ≥ sqrt (4 pq)

The solution bank added the line p and q are both positive
so p > 0 and q > 0
Therefore, p + q > 0

Why is this necessary? Doesn't the line "...(p-q)^2 and any number squared is ≥ 0 " eliminate the need for it, since it shows that 4qr is the value that (p +q)^2 is equal or greater than?
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#2
*bump* whoops
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1 month ago
#3
(Original post by EmRep13)
I did (p + q)^2 = p^2 + 2pq + q^2 g ≥ 4pq
so p^2 -2qr + q^2 ≥ 0 which is true since it's equal to (p-q)^2 and any number squared is ≥ 0.
Since (p - q)^2 + 4pq = (p + q)^2 and (p - q)^2 ≥ 0
(p + q)^2 ≥ 4pq so
p + q ≥ sqrt (4 pq)

The solution bank added the line p and q are both positive
so p > 0 and q > 0
Therefore, p + q > 0

Why is this necessary? Doesn't the line "...(p-q)^2 and any number squared is ≥ 0 " eliminate the need for it, since it shows that 4qr is the value that (p +q)^2 is equal or greater than?
Nowhere in your proof do you use that p & q are positive....
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#4
(Original post by zetamcfc)
Nowhere in your proof do you use that p & q are positive....
I don't see why you'd need to? Other than in the last bit just to state that it only applies if p + q are positive? But is that grounds to do so? Do you need to use it if the question states?
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1 month ago
#5
(Original post by EmRep13)
*bump* whoops
When they square rooted both sides of the inequality (p+q)^2 >= 4pq they used the fact that p,q are both positive (a,b>0 a^2>b^2 <==> a>b)) so you need to state this EmRep13
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1 month ago
#6
(Original post by EmRep13)
I don't see why you'd need to? Other than in the last bit just to state that it only applies if p + q are positive? But is that grounds to do so? Do you need to use it if the question states? but , the problem comes when you go from to . (some constraint on , is required for this inequality to even make sense)
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