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Cum se demonstreaza (x+y)^2 >= 4xy?

Răspuns :

x^2+y^2+2xy>=4xy
x^2+y^2>=2xy
x^2+y^2-2xy>=0
(x-y)^2>=0 adevarat
(x+y)²=4xy
x²+2xy+y²=4xy
x²-2xy+y²=0
(x-y)²=0
=> (x+y)²=4xy ⇔ x-y=0