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huynhbachkhoa23

$S=(x+y)^2.\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{(x+y)^2}{2xy}$
Áp dụng Cauchy-Schwarz: $\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge \dfrac{4}{(x+y)^2}$ nên $S\ge 4+\dfrac{(x+y)^2}{2xy}\ge 6$
 
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