cho 3 so a b c thoa man a+b+c=3/2 chung minh a^2+b^2+c^2>=3/4 01/11/2021 Bởi Elliana cho 3 so a b c thoa man a+b+c=3/2 chung minh a^2+b^2+c^2>=3/4
Ta có: $(a-b)^2+(b-c)^2+(c-a)^2$$\geq$$0$ ⇔$a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2$$\geq$$0$ ⇔$2a^2+2b^2+2c^2-2ab-2bc-2ca$$\geq$$0$ ⇔$2a^2+2b^2+2c^2$$\geq$$2ab+2bc+2ca$ ⇔$3(a^2+b^2+c^2)$$\geq$$a^2+b^2+c^2+2ab+2bc+2ca$ ⇔$3(a^2+b^2+c^2)$$\geq$$(a+b+c)^2=$$($$\dfrac{3}{2}$$)^2=$$\dfrac{9}{4}$ ⇔$(a^2+b^2+c^2)$$\geq$$\dfrac{3}{4}$ Dấu = xảy ra ⇔$a=b=c=$$\dfrac{1}{2}$ Bình luận
Ta có:
$(a-b)^2+(b-c)^2+(c-a)^2$$\geq$$0$
⇔$a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2$$\geq$$0$
⇔$2a^2+2b^2+2c^2-2ab-2bc-2ca$$\geq$$0$
⇔$2a^2+2b^2+2c^2$$\geq$$2ab+2bc+2ca$
⇔$3(a^2+b^2+c^2)$$\geq$$a^2+b^2+c^2+2ab+2bc+2ca$
⇔$3(a^2+b^2+c^2)$$\geq$$(a+b+c)^2=$$($$\dfrac{3}{2}$$)^2=$$\dfrac{9}{4}$
⇔$(a^2+b^2+c^2)$$\geq$$\dfrac{3}{4}$
Dấu = xảy ra ⇔$a=b=c=$$\dfrac{1}{2}$