Hello, could someone help with proving this inequality:
$\displaystyle a^2+b^2+1 \ge ab + a + b$
Thanks in advance
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 $\displaystyle b = ka$ gives rise a quadratic inequality on $\displaystyle a$ where coefficients are quadratic polynomials of $\displaystyle k$. By considering the leading coefficient and the discriminant, one can show that the inequality holds.
You could also utilise the Arithmetic Mean -- Geometric Mean Inequality
$\displaystyle \displaystyle\frac{a+b}{2}\ge\sqrt{ab}$
Therefore
$\displaystyle \displaystyle\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}$
$\displaystyle \displaystyle\frac{a^2+1}{2}\ge\sqrt{a^2}$
$\displaystyle \displaystyle\frac{b^2+1}{2}\ge\sqrt{b^2}$
Hence, in summing these
$\displaystyle a^2+b^2+1\ge\ |ab|+|a|+|b|\Rightarrow\ a^2+b^2+1\ge\ ab+a+b$