# to prove $x^2 + y^2+1\ge xy + y + x$

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

$$x$$ 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??

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

Alternatively,

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

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 $$f(1,1)=0\geq 0$$.