proof of a^2 + ab + b^2 >0; a,b, both are not zero, both are real

Question

Show that for any reals a,b, both of which are not zero, it holds true that a^2 + ab + b^2 >0

"Hint": $$a^2+ab+b^2=\dfrac{(a+b)^2+a^2+b^2}{2}.$$ – CIJ Oct 13 '15 at 04:53 — CIJ, Oct 13 '15 at 04:53
If an answer satisfies you, could you mark it as such. – Jean-François Gagnon Oct 13 '15 at 17:19 — Jean-François Gagnon, Oct 13 '15 at 17:19

score 2 · Answer 1 · answered Oct 13 '15 at 04:58

2

$$(a+b)^2=a^2+2ab+b^2$$ So you want to show $$(a+b)^2-ab>0$$ Or $$(a+b)^2>ab$$ If $a$ and $b$ are not of the same sign you are done. If not, consider the rectangle of sides $a$ and $b$ and see that it fits inside the square of side $a+b$.

answered Oct 13 '15 at 04:58

Jean-François Gagnon

1,100

score 1 · Answer 2 · answered Oct 13 '15 at 04:52

1

$$a^2+ab+b^2=\dfrac{(2a+b)^2+(\sqrt3b)^2}4\ge\dfrac{(\sqrt3b)^2}4$$

answered Oct 13 '15 at 04:52

lab bhattacharjee

279,016

proof of a^2 + ab + b^2 >0; a,b, both are not zero, both are real

2 Answers2