proving an inequality related to $AM\ge GM$ $$a^2+ab+b^2\ge 3(a+b-1)$$
$a,b$ are real numbers
using $AM\ge GM$
I proved that 
$$a^2+b^2+ab\ge 3ab$$
$$(a^2+b^2+ab)/3\ge 3ab$$
how do I prove that $$3ab\ge 3(a+b-1)$$
if I'm able to prove the above inequality then i'll be done
 A: Hint:  Let $a=x+1$ and $b=y+1$ (in the original inequality).
A: This type of inequality can be proved in the similar way like this:
$$ a^2 + ab + b^2 - 3(a + b - 1) = a^2 + ab + b^2 - 3a - 3b + 3 $$
$$ = a^2 - 2a + 1 + b^2 - 2b + 1 + ab -a -b + 1 = (a - 1)^2 + (b-1)^2 +(a-1)(b-1) $$ $$\ge 2|a-1||b-1| - (a -1)(b-1) \ge 0$$
The last one comes from AM-GM inequality.
A: Follow @BarryCipra's hint of letting $a=x+1$ and $b=y+1$:
$$(x+1)^2+(x+1)(y+1)+(y+1)^2 \geq 3((x+1)+(y+1)-1)$$
Expand the left side and simplify the right side:
$$x^2+2x+1+xy+x+y+1+y^2+2y+1 \geq 3(x+y+1)$$
Simplify the left side and distribute the $3$ on the right side:
$$x^2+xy+y^2+3x+3y+3 \geq 3x+3y+3$$
Subtract both sides by $3x+3y+3$:
$$x^2+xy+y^2 \geq 0$$
Now, to prove this. we need to make some changes. We know that $xy \geq \lvert xy \rvert$. We also know that $\lvert xy \rvert \geq -\lvert xy \rvert$ because $\lvert xy \rvert$ is non-negative and non-negative numbers are always greater than or equal to their negatives. Finally, we have $-\lvert xy \rvert \geq -2\lvert xy \rvert$ because a non-positive number added to itself is always less than or equal to the original (e.g. $-2+(-2) \leq -2$, $-1+(-1) \leq -1$, $0+0 \leq 0$). Therefore, by the Transitive Property of Equality, we have $xy \geq -2\lvert xy \rvert$ and by adding $x^2+y^2$ by both sides of the inequality, we have $x^2+xy+y^2 \geq x^2-2\lvert xy \rvert+y^2$. This means we only need to prove the latter is non-negative to prove the former is non-negative, so we've reduced the problem to:
$$x^2-2\lvert xy \rvert+y^2 \geq 0$$
Now, we know that $x^2=\lvert x \rvert^2$, $y^2=\lvert y\rvert^2$, and $\lvert xy \rvert=\lvert x \rvert \lvert y \rvert$, so we can substitute to get everything in terms of $\lvert x \rvert$ and $\lvert y \rvert$:
$$\lvert x \rvert^2-2\lvert x \rvert \lvert y \rvert+\lvert y \rvert^2 \geq 0$$
Clearly, this is the same as $(\lvert x \rvert - \lvert y \rvert)^2$:
$$(\lvert x \rvert - \lvert y \rvert)^2 \geq 0$$
This is obviously true because the square of a real number is always non-negative. Thus, by proving this is non-negative, we have proven that $x^2+xy+y^2$ is non-negative by our previous argument, which proves the whole inequality.
