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This question already has an answer here:

Given $x,y \in \Bbb{R}$, show that:$$x^2+y^2+1\ge xy+y+x $$ I tried using the fact that $x^2+y^2 \ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $\frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1\ge x+y$

This inequality seems very easy, I'm feeling dumb for not having solved it yet

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marked as duplicate by Martin R, Davide Giraudo, Did, kjetil b halvorsen, RRL Dec 16 '18 at 14:13

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  • $\begingroup$ Fix $y$, differentiate w.r.t. $x$, etc $\endgroup$ – mathworker21 Dec 16 '18 at 10:58
  • $\begingroup$ @MartinR I just didn't see them being equivalent, I well new that one inequality $\endgroup$ – Spasoje Durovic Dec 16 '18 at 11:07
  • $\begingroup$ @Spasoje Durovic Substitute $c=1$. $\endgroup$ – Michael Rozenberg Dec 16 '18 at 11:11
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Since $$a^2-2ab+b^2 = (a-b)^2\geq 0\implies a^2+b^2\geq 2ab$$ we have $$ x^2+y^2\geq 2xy$$ $$x^2+1\geq 2x$$ $$y^2+1\geq 2y$$ Now add all these...

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Hint: Use that $$x^2+y^2+z^2\geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$ this is $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$

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We need to prove that $$y^2-(x+1)y+x^2-x+1\geq0,$$ for which it's enough to prove that $$(x+1)^2-4(x^2-x+1)\leq0$$ or $$(x-1)^2\geq0.$$

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