$$x^2 + y^2+1\ge xy + y + x$$

$x$ and $ y$ belong to all real numbers

my attempt

$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $

let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$

$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$

$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality

$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)\ge3xy+(x+y)$

hence $\Rightarrow x^2 + y^2+2xy+1\ge 3xy+x+y$

are the steps correct and is there any other better way??

  • $\begingroup$ "But $\frac{(x+y)^2}{4} \ge xy $" - why? $\endgroup$ – Dietrich Burde Apr 23 at 8:53
  • $\begingroup$ Type/bounds of x,y? $\endgroup$ – NoChance Apr 23 at 8:55
  • $\begingroup$ @DietrichBurde That is because $(x-y)^{2} \geq 0$. $\endgroup$ – Kabo Murphy Apr 23 at 8:57
  • $\begingroup$ @KaviRamaMurthy Yes, I know. But it should be mentioned in the solution. $\endgroup$ – Dietrich Burde Apr 23 at 9:02
  • $\begingroup$ @DietrichBurde by inequality of arithmetic and geometric mean, followed by squaring on both sides $\endgroup$ – Snmohith Raju Apr 23 at 9:18

Let $c=x^2+y^2+1-(xy+x+y)$

$\iff x^2-x(1+y)+1-y+y^2-c=0$

As $x$ is real, the discriminant must be $\ge0$

i.e., $$(1+y)^2\ge4(1-y+y^2-c)\iff4c\ge3(1-y)^2$$ which is $\ge0$ for real $y$


$$ x^2-x(1+y)+1-y+y^2=\left(x-\dfrac{1+y}2\right)^2+\dfrac{3(1-y)^2}4$$


Prove $$f(x,y)=x^2+y^2+1-x-xy-y\geq 0.$$ $$f_x=2x-1-y$$ $$f_y=2y-1-x$$ $$f_x=0\implies y=2x-1$$ $$f_y=0\implies x=2y-1$$ $$y=2(2y-1)-1=4y-3\implies y=1$$ $$x=2(2x-1)-1=4x-3\implies x=1$$

There's a stationary point at $(1,1)$. $$f_{xx}=2,f_{xy}=f_{yx}=-1,f_{yy}=2.$$

Since $$f_{xx}f_{yy}-f_{xy}^2=4-(-1)^2=3>0\text{ and }f_{xx}=2>0$$ then that stationary point is a minimum. Furthermore $f(1,1)=0\geq 0$.


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