Đáp án: $ P\ge\dfrac{(\sqrt{10}+\sqrt2)^2}{3+\sqrt5}$
Giải thích các bước giải:
Ta có:
$P=\dfrac1a+\dfrac2b$
$\to P=\dfrac{10}{10a}+\dfrac{2}{b}$
$\to P=\dfrac{(\sqrt{10})^2}{10a}+\dfrac{(\sqrt2)^2}{b}$
$\to P\ge\dfrac{(\sqrt{10}+\sqrt2)^2}{10a+b}$
$\to P\ge\dfrac{(\sqrt{10}+\sqrt2)^2}{3+\sqrt5}$
Dấu = xảy ra khi
$\begin{cases}\dfrac{\sqrt{10}}{10a}=\dfrac{\sqrt2}{b}\\ 10a+b=3+\sqrt5\end{cases}$
$\to \dfrac{\sqrt{10}}{10a}=\dfrac{\sqrt2}{b}=\dfrac{\sqrt{10}+\sqrt2}{10a+b}$
$\to \dfrac{\sqrt{10}}{10a}=\dfrac{\sqrt2}{b}=\dfrac{\sqrt{10}+\sqrt2}{3+\sqrt5}$
$\to a=\dfrac{5+\sqrt{5}}{20}, b=\dfrac{\sqrt{5}+1}{2}$
