Question

Cho \(\left\{{}\begin{matrix}a,b>0\\a+b=1\end{matrix}\right.\) . Tìm Min :

P= \(\dfrac{18}{a^2+b^2}+\dfrac{5}{ab}\)

Answer 1

\(P=\dfrac{18}{a^2+b^2}+\dfrac{10}{2ab}\ge\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)^2}{a^2+b^2+2ab}=28+12\sqrt{5}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b=1\\\dfrac{a^2+b^2}{3}=\dfrac{2ab}{\sqrt{5}}\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(\dfrac{5-\sqrt{5}}{4};\dfrac{\sqrt{5}-1}{4}\right)\) và hoán vị