Hi
If \( \displaystyle p,q,r \) are roots of \( x^3+3x^2-1=0 \).
Then \( p^2q+q^2r+r^2p= \)
You can define:
\( f(x,y,z)=x^2y+y^2z+z^2x \)
\( g(p,q,r)=x^2z+y^2z+r^2q \)
They are not symmetric polynomials but their sum and product are. They can be expressed in terms of elementary symmetric polynomials:
\( s_1(x,y,z)=x+y+z,\quad s_2(x,y,z)=xy+xz+yx,\quad s_3(x,y,z)=xyz \)
\( f+g=-3s_3+s_1\cdot s_2 \) (*)
\( f\cdot g=9s_3^2 + s_3\cdot s_1^3 - 6 s_3s_1s_2+s_2^3 \)(**)
Now, by Vietta-Cardano relations of the roots of a polynomial, if \( p,q,r \) are the roots of \( x^3+3x^2-1=0 \).
\( s_1(p,q,r)=-3,\quad s_2(p,q,r)=0,\quad s_3(p,q,r)=1 \)
From this and (*),(**):
\( f(p,q,r)+g(p,q,r)=-3 \) (1)
\( f(p,q,r)g(p,q,r)=-18 \) (2)
Solving the system:
\( \{f(p,q,r),g(p,q,r)\}=\{3,-6\} \)
Now, we only need to check that \( f(p,q,r)>g(p,q,r) \).
Best regards.
