Đáp án: $ P\ge \sqrt{3}$
Giải thích các bước giải:
Ta có:
$P=a^2+b^2+\dfrac{1}{a^2}+\dfrac{b}{a}$
$\to P=b^2+2\cdot b\cdot \dfrac1{2a}+\left(\dfrac1{2a}\right)^2+\dfrac3{4a^2}+a^2$
$\to P=\left(b+\dfrac1{2a}\right)^2+\dfrac3{4a^2}+a^2$
$\to P\ge 0+2\sqrt{\dfrac3{4a^2}\cdot a^2}$
$\to P\ge \sqrt{3}$
Dấu = xảy ra khi:
$\begin{cases}\dfrac3{4a^2}=a^2\\ b+\dfrac1{2a}=0\end{cases}$
$\to \left(a,b\right)\in\{\left(\sqrt{\dfrac{\sqrt{3}}{2}},-\dfrac{1}{2\sqrt{\dfrac{\sqrt{3}}{2}}}\right),\left(-\sqrt{\dfrac{\sqrt{3}}{2}},\dfrac{1}{2\sqrt{\dfrac{\sqrt{3}}{2}}}\right)\}$
