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

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

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2 Answers

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$$(a+b)^2=a^2+2ab+b^2$$ 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 $a+b$.

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

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