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Question Number 181099 by Ari last updated on 21/Nov/22
  prove that for every   positivenumber p e q wee  have:  p+q≥(√(4pq))
provethatforeverypositivenumberpeqweehave:p+q4pq
Answered by Agnibhoo98 last updated on 21/Nov/22
When p and q are positive numbers  (p − q)^2  ≥ 0  or p^2  − 2pq + q^2  ≥ 0  or p^2  − 2pq + 4pq + q^2  ≥ 4pq (Adding 4pq both side)  or p^2  + 2pq + q^2  ≥ 4pq  or (p + q)^2  ≥ 4pq  or p + q ≥ (√(4pq))     (Proved)
Whenpandqarepositivenumbers(pq)20orp22pq+q20orp22pq+4pq+q24pq(Adding4pqbothside)orp2+2pq+q24pqor(p+q)24pqorp+q4pq(Proved)
Commented by Ari last updated on 21/Nov/22
Thanks Mr∫!
ThanksMr!
Commented by Agnibhoo98 last updated on 22/Nov/22
Another way  According to AM ≥ GM method :  ((p + q)/2) ≥ (√(pq))  or p + q ≥ 2(√(pq))  or p + q ≥ (√(4pq)) (Proved)
AnotherwayAccordingtoAMGMmethod:p+q2pqorp+q2pqorp+q4pq(Proved)
Answered by SEKRET last updated on 22/Nov/22
((√(p ))  − (√q) )^2  ≥ 0    p+q≥ (√(4pq))
(pq)20p+q4pq

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