Prove the inequality: $a^2+b^2+1≥ab+a+b$

Question

Prove the inequality.

$a^2+b^2+1≥ab+a+b$

I try so many methods, But I have not been successful in any way.Because, I can not find "hint".

Answer 1

$$a^2+b^2+1-ab-a-b=\frac{1}{2}((a-b)^2+(a-1)^2+(b-1)^2)\geq0$$

Answer 2

This is the $c=1$ case of $$a^2+b^2+c^2\ge ab+ac+bc.$$ One only needs to prove this for $a$, $b$, $c\ge0$.

Answer 3

If $b=1$ it becomes $a^2-2a+1\ge 0$ which is true. assume $b\ne 1$.

It is equivalent to prove that $$a^2-a (b+1)+b^2+1-b>0$$

$$\Delta=b^2+2b+1-4b^2-4+4b $$ $$=-3(b^2-2b+1)=-3 (b-1)^2<0$$ Done.

Answer 4

All of the quantities on the left are $+ve$ ... \begin{eqnarray*} (a-b)^2+(a-1)^2+(b-1)^2 \geq 0 \end{eqnarray*} divide by $2$ & rearrange ... the inequality follows.

Answer 5

Move all to left: $$a^2+b^2+1-ab-a-b\ge 0$$

Find the minimum value of $f(a,b)=a^2+b^2+1-ab-a-b$ using partial differentiation:

$f_a=2a-b-1=0$

$f_b=2b-a-1=0$

$a=b=1$

$f_{aa}=2>0, f_{bb}=2>0, f_{aa}f_{bb}-f_{ab}^2=3>0 \Rightarrow f(1,1)=0$ is minimum.