Question

Cho \(a,b>0:\dfrac{a}{1+a}+\dfrac{b^3}{b+1}=1\). Tìm max: P=\(ab^3\)

Answer 1

\(\dfrac{a}{1+a}+\dfrac{b^3}{1+b}=1\Rightarrow\dfrac{a}{1+a}=1-\dfrac{b^3}{1+b}=\dfrac{1+b-b^3}{1+b}\)

\(\Rightarrow a\left(1+b\right)=\left(1+a\right)\left(1+b-b^3\right)\)

\(\Rightarrow a\left(1+b\right)=1+b-b^3+a\left(1+b-b^3\right)\)

\(\Rightarrow ab^3=1+b-b^3=1+3.\dfrac{1}{\sqrt[]{3}}.\dfrac{1}{\sqrt[]{3}}b-b^3\le1+\left(\dfrac{1}{\sqrt[]{3}}\right)^3+\left(\dfrac{1}{\sqrt[]{3}}\right)^3+b^3-b^3=\dfrac{9+2\sqrt[]{3}}{9}\)

\(P_{max}=\dfrac{9+2\sqrt[]{3}}{9}\) khi \(\left(a;b\right)=\left(2+3\sqrt[]{3};\dfrac{1}{\sqrt[]{3}}\right)\)