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Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+1 \geq x+xy$.

It seems like I should apply AM-GM here, but I don't see how it helps.

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    $\begingroup$ how is $z$ related to this problem $\endgroup$ – ThePortakal Dec 26 '15 at 23:20
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    $\begingroup$ What is the point of $z$? $\endgroup$ – mysatellite Dec 26 '15 at 23:21
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    $\begingroup$ @JacobWillis, in five days you have asked 12 questions, but have accepted a single answer. This kind of behaviour is discouraged here. It is normal not to accept answers only if they are of poor quality, but from what I have seen, you have received some very good ones. Please review your questions and, where the given answers helped you solve your problem, think about accepting them. $\endgroup$ – Alex M. Dec 27 '15 at 10:47
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We have: $$ \frac{x^2+1}{2}≥|x| \\ \frac{x^2+y^2}{2}≥|xy| \\ \frac{y^2+1}{2}≥|y| $$ By AM_GM. Thus, adding yields: $$ x^2+y^2+1≥|xy|+|x|+|y|≥xy+x $$

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We have a stronger inequality: $$x^2+y^2+1-x-xy=\frac34\left(x-\frac23\right)^2+\left(\frac12x-y\right)^2+\frac23\geq\frac23\,.$$ The equality holds iff $x=\frac23$, $y=\frac13$, and $z$ is an arbitrary object, whatever it is supposed to be.

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No need for the AM-GM inequality here: rewrite this as $$\frac {x^2} 2 - x + \frac 1 2 + \frac {x^2} 2 - xy + \frac {y^2} 2 + \frac {y^2} 2 + \frac 1 2 \ge 0$$ which is trivially true, because the left-hand side is a sum of clearly positive numbers: $$\left( \frac x {\sqrt 2} - \frac 1 {\sqrt 2} \right) ^2 + \left( \frac x {\sqrt 2} - \frac y {\sqrt 2} \right) ^2 + \frac {y^2} 2 + \frac 1 2 .$$

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If $x^2+y^2+1-x-x y +d^2=0$ then $x^2+x(-1-y)+(y^2+1+d^2)=0$ which, seen as a quadratic equation in $x$, must have a non-negative discriminant. That is $(1+y)^2-4(y^2+1+d^2)\geq 0$. This is easily seen to be impossible even for $d=0$.

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