\(P=\dfrac{4ab}{\left(a+b\right)^2}+\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{4ab}{\left(a+b\right)^2}+\dfrac{a^2+b^2}{2ab}+\dfrac{a^2+b^2}{2ab}\)
-Áp dụng BĐT \(\dfrac{A+C}{B+C}\ge1\) với \(\dfrac{A}{B}\ge1\) và \(C>0\), ta có:
\(P=\dfrac{4ab}{\left(a+b\right)^2}+\dfrac{a^2+b^2}{2ab}+\dfrac{a^2+b^2}{2ab}\ge\dfrac{4ab}{\left(a+b\right)^2}+\dfrac{2a^2+2b^2}{\left(a+b\right)^2}+\dfrac{a^2+b^2}{2ab}=\dfrac{2\left(a+b\right)^2}{\left(a+b\right)^2}+\dfrac{a^2+b^2}{2ab}=2+\dfrac{a^2+b^2}{2ab}\)
-Áp dụng BĐT AM-GM ta có:
\(P\ge2+\dfrac{a^2+b^2}{2ab}\ge2+\dfrac{2ab}{2ab}=2+1+3\)
\(P=3\Leftrightarrow a=b\)
-Vậy \(P_{min}=3\)
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