Question

Answer 1

Ta có:

\(a^3+a^3+1\ge3a^2\Rightarrow a^3\ge\dfrac{3a^2-1}{2}\)

\(b^2+c^2\ge\dfrac{1}{2}\left(b+c\right)^2=\dfrac{1}{2}\left(4-a\right)^2\)

\(\Rightarrow P\ge\dfrac{3a^2-1}{2}+\dfrac{1}{2}\left(4-a\right)^2=2\left(a-1\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\)

\(P_{min}=\dfrac{11}{2}\) khi \(\left(a;b;c\right)=\left(1;\dfrac{3}{2};\dfrac{3}{2}\right)\)