Cho biểu thức P = a^4 + b^4 - ab với a, b là các số thực thỏa mãn a^2 + b^2 + ab = 3

Ta có $a^2+b^2+ab=3\iff a^2+b^2=3-ab$ thay vào P ta được.

$$\begin{gathered} P=a^4+b^4-ab={\left(a^2+b^2\right)}^2-2a^2{b}^2-ab \\ ={\left(3-ab\right)}^2-2a^2{b}^2-ab \\ =9-6ab+a^2{b}^2-2a^2{b}^2-ab \\ =9-7ab-a^2{b}^2 \\ =-\left[{\left(ab\right)}^2+2.ab.\frac{7}{2}+\frac{49}{4}\right]+\frac{49}{4}+9 \\ =-{\left(ab+\frac{7}{2}\right)}^2+\frac{85}{4} \end{gathered}$$

Vì $a^2+b^2=3-ab$, mà ${\left(a+b\right)}^2\ge 0\iff a^2+b^2\ge -2ab\Rightarrow 3-ab\ge -2ab\iff ab\ge -3$. (1)

Và ${\left(a-b\right)}^2\ge 0\iff a^2+b^2\ge 2ab\Rightarrow 3-ab\ge 2ab\iff ab\le 1$. (2)

Từ (1) và (2) suy ra $-3\le ab\le 1\iff -3+\frac{7}{2}\le ab+\frac{7}{2}\le \frac{7}{2}+1\iff \frac{1}{2}\le ab+\frac{7}{2}\le \frac{9}{2}$

$$\begin{gathered} \iff \frac{1}{4}\le {\left(ab+\frac{7}{2}\right)}^2\le \frac{81}{4} \\ \iff -\frac{81}{4}\le -{\left(ab+\frac{7}{2}\right)}^2\le -\frac{1}{4} \\ \iff -\frac{81}{4}+\frac{85}{4}\le -{\left(ab+\frac{7}{2}\right)}^2+\frac{85}{4}\le -\frac{1}{4}+\frac{85}{4} \\ \iff 1\le -{\left(ab+\frac{7}{2}\right)}^2+\frac{85}{4}\le 21 \end{gathered}$$ 

Vậy Max P = 21. Dấu = xảy ra khi $\left\{\begin{array}{l}ab=-3\\ a^2+b^2=6\end{array}\right.$$\iff \left\{\begin{array}{l}a=\sqrt{3}\\ b=-\sqrt{3}\end{array}\right.v\left\{\begin{array}{l}b=\sqrt{3}\\ a=-\sqrt{3}\end{array}\right.$.

Min P = 1. Dấu = xảy ra khi $\left\{\begin{array}{l}ab=1\\ a^2+b^2=2\end{array}\right.\iff \left\{\begin{array}{l}a=1\\ b=1\end{array}\right.$ hoặc $\left\{\begin{array}{l}a=-1\\ b=-1\end{array}\right.$.