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Prove that $a^2+ab+b^2\geq 0$ for all $a,b$. My attempt

I supossed by contradiction that exist $a,b$ such that $a^2+b^2<-ab$ then $a^2+2ab+b^2\leq ab$ implies that $(a+b)^2\leq ab$ therefore $a+b<\sqrt{ab}$ but $\sqrt{ab}\leq \frac{a+b}{2}$ then $a+b<\frac{a+b}{2}$ contradiction.

Is fine my proof? Or exist other form to show. Thank you

weymar andres
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  • @user The proof is valid only for $a \geq 0 $and $b \geq 0$. If $a<0$ and $b<0$ then $ab \geq 0$ is true but the step $\sqrt{ab} \leq \frac{a+b}{2}$ (AM-GM) is not true. – aschepler Sep 19 '22 at 14:31
  • thank you @ascheper!! – weymar andres Sep 19 '22 at 14:33
  • @aschepler Yes of course! thanks – user Sep 19 '22 at 14:34
  • When you assume $(a+b)^2 \le ab \implies a+b \le \sqrt {ab}$ you are assuming that $ab \ge 0$ and that $a + b\ge 0$. In assume $\sqrt{ab}\le \frac {a+b}2$ you are assuming $a,b > 0$. None of these are given. But you are on the way. Note: $0 \le a^2 + b^2 <-ab \implies ab< 0$. Then $a^2+2ab + b^2 \le ab\implies (a+b)^2 = a^2 + 2ab + b^2 \le ab < 0$ which is a contradiction. – fleablood Sep 19 '22 at 19:14

2 Answers2

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$$a^2+b^2+ab=\frac{1}{2}[(a+b)^2+(a-b)^2]+\frac{1}{4}[(a+b)^2-(a-b)^2]=\frac{3}{4}(a+b)^2+\frac{1}{4}(a-b)^2 \ge 0$$

Z Ahmed
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$a^2+ab+b^2= (a+b)^2-ab.$ If $a$ and $b$ have opposite sign, or at least one is $0,$ the result is obvious since the right-hand side will be nonnegative. If $a$ and $b$ have the same sign, there is no harm assuming both are positive. But then the left-hand side of the equality is positive, and you are done.

Or, notice $|ab| \leq \mathrm{max}\{a^2,b^2\},$ so either $a^2+ab$ or $b^2+ab$ is nonnegative (even if $ab$ is negative) and you're done.

Chris Leary
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