1
$\begingroup$

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

$x$ 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??

$\endgroup$
  • $\begingroup$ "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
3
$\begingroup$

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,

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.