To prove that a2−ab+b2≥1a^2 - ab + b^2 \geq 1a2−ab+b2≥1 when (a > 0) and (b > 0), we can rewrite the inequality as:
a2−ab+b2−1≥0a^2 - ab + b^2 - 1 \geq 0a2−ab+b2−1≥0
Now, let's consider the expression a2−ab+b2−1a^2 - ab + b^2 - 1a2−ab+b2−1 as a quadratic in terms of aaa. The discriminant of this quadratic expression is given by:
Since (a > 0) and (b > 0), we know that (b > 0), which implies that (-3b^2 < 0). Therefore, (\Delta < 0), which means that the quadratic expression a2−ab+b2−1a^2 - ab + b^2 - 1a2−ab+b2−1 will always be greater than or equal to 0 when (a > 0) and (b > 0).
Therefore, we can conclude that a2−ab+b2≥1a^2 - ab + b^2 \geq 1a2−ab+b2≥1 when (a > 0) and (b > 0).
To prove that a2−ab+b2≥1a^2 - ab + b^2 \geq 1a2−ab+b2≥1 when (a > 0) and (b > 0), we can rewrite the inequality as:
a2−ab+b2−1≥0a^2 - ab + b^2 - 1 \geq 0a2−ab+b2−1≥0
Now, let's consider the expression a2−ab+b2−1a^2 - ab + b^2 - 1a2−ab+b2−1 as a quadratic in terms of aaa. The discriminant of this quadratic expression is given by:
Δ=(−b)2−4(1)(b2−1)=b2−4b2+4=−3b2+4\Delta = (-b)^2 - 4(1)(b^2 - 1) = b^2 - 4b^2 + 4 = -3b^2 + 4Δ=(−b)2−4(1)(b2−1)=b2−4b2+4=−3b2+4
Since (a > 0) and (b > 0), we know that (b > 0), which implies that (-3b^2 < 0). Therefore, (\Delta < 0), which means that the quadratic expression a2−ab+b2−1a^2 - ab + b^2 - 1a2−ab+b2−1 will always be greater than or equal to 0 when (a > 0) and (b > 0).
Therefore, we can conclude that a2−ab+b2≥1a^2 - ab + b^2 \geq 1a2−ab+b2≥1 when (a > 0) and (b > 0).