a/ Ta có:
\(\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\le\frac{a+b-c+b+c-a}{2}=b\left(1\right)\)
Tương tự ta có:
\(\hept{\begin{cases}\sqrt{\left(a+b-c\right)\left(c+a-b\right)}\le a\left(2\right)\\\sqrt{\left(b+c-a\right)\left(c+a-b\right)}\le c\left(3\right)\end{cases}}\)
Lấy (1), (2), (3) nhân vế theo vế ta được
\(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)
b/
\(a^3+b^3+c^3+2abc< a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)\)
\(\Leftrightarrow\left[ab^2+ac^2-a^3\right]+\left[ba^2+bc^2-b^3\right]+\left[ca^2+cb^2-c^3\right]>2abc\)
\(\Leftrightarrow\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{c^2+a^2-b^2}{2ca}+\dfrac{a^2+b^2-c^2}{2ab}-1>0\)
\(\Leftrightarrow\dfrac{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}{2abc}>0\) (đúng)
Vậy ta có ĐPCM