a bình + 2a + 2b bình +b/2 +5/4 lớn hơn hoặc bằng 0 22/08/2021 Bởi Ayla a bình + 2a + 2b bình +b/2 +5/4 lớn hơn hoặc bằng 0
Ta có $a^2 + 2a + 2b^2 + \dfrac{b}{2} + \dfrac{5}{4} = (a^2 + 2a + 1) + (2b^2 + \dfrac{b}{2} + \dfrac{1}{32} ) + \dfrac{7}{32}$ $= (a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 + \dfrac{7}{32}$ Ta có $(a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 \geq 0$ $<-> (a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 + \dfrac{7}{32} \geq \dfrac{7}{32} > 0$ Bình luận
Ta có
$a^2 + 2a + 2b^2 + \dfrac{b}{2} + \dfrac{5}{4} = (a^2 + 2a + 1) + (2b^2 + \dfrac{b}{2} + \dfrac{1}{32} ) + \dfrac{7}{32}$
$= (a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 + \dfrac{7}{32}$
Ta có
$(a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 \geq 0$
$<-> (a+1)^2 + (b\sqrt{2} + \dfrac{1}{4\sqrt{2}})^2 + \dfrac{7}{32} \geq \dfrac{7}{32} > 0$