# Simple two variable am-gm inequality [duplicate]

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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

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

• Fix $y$, differentiate w.r.t. $x$, etc – mathworker21 Dec 16 '18 at 10:58
• @Spasoje Durovic Substitute $c=1$. – Michael Rozenberg Dec 16 '18 at 11:11
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...
Hint: Use that $$x^2+y^2+z^2\geq xy+yz+zx$$ holds for all real numbers $$x,y,z$$ this is $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$
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 $$(x-1)^2\geq0.$$