Question

Cho a,b,c>0 và a+b+c=6. Chứng minh rằng: 
5a^3+b^2+c^2>=27/2

 

Answer 1

Nếu đề là \(a+b+c=6\) thì \(5a^3+b^2+c^2\ge\dfrac{783}{50}\) mới là BĐT hợp lý

Answer 2

Đặt \(P=5a^3+b^2+c^2\ge5a^3+\dfrac{1}{2}\left(b+c\right)^2=5a^3+\dfrac{1}{2}\left(6-a\right)^2\)

\(\Rightarrow P\ge5a^3+\dfrac{1}{2}a^2-6a+18=5a^3+\dfrac{1}{2}a^2-6a+\dfrac{117}{50}+\dfrac{783}{50}\)

\(\Rightarrow P\ge\dfrac{1}{50}\left(5a-3\right)^2\left(10a+13\right)+\dfrac{783}{50}\)

Do \(\dfrac{1}{50}\left(5a-3\right)^2\left(10a+13\right)\ge0;\forall a>0\)

\(\Rightarrow P\ge\dfrac{783}{50}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{5};\dfrac{27}{10};\dfrac{27}{10}\right)\)