1. ## Prove an inequality

Hello, could someone help with proving this inequality:

$a^2+b^2+1 \ge ab + a + b$

2. One way is to consider a line b = k * a for some number k and to prove that the inequality holds for all points (a, b) on this line. (For completeness, one should consider the line a = 0, but the inequality is symmetric w.r.t. a and b, so this is similar to k = 0.)

The substitution $b = ka$ gives rise a quadratic inequality on $a$ where coefficients are quadratic polynomials of $k$. By considering the leading coefficient and the discriminant, one can show that the inequality holds.

3. Originally Posted by mat1990
Hello, could someone help with proving this inequality:

$a^2+b^2+1 \ge ab + a + b$

You could also utilise the Arithmetic Mean -- Geometric Mean Inequality

$\displaystyle\frac{a+b}{2}\ge\sqrt{ab}$

Therefore

$\displaystyle\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}$

$\displaystyle\frac{a^2+1}{2}\ge\sqrt{a^2}$

$\displaystyle\frac{b^2+1}{2}\ge\sqrt{b^2}$

Hence, in summing these

$a^2+b^2+1\ge\ |ab|+|a|+|b|\Rightarrow\ a^2+b^2+1\ge\ ab+a+b$

4. Originally Posted by mat1990
Hello, could someone help with proving this inequality:

$a^2+b^2+1 \ge ab + a + b$

$(a-b)^2+(a-1)^2+(b-1)^2\ge\ 0$
$\Rightarrow\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1\ge\ 0$
$2\left(a^2+b^2+1\right)\ge\ 2(ab+a+b)$