$$2s_1^2-4s_2+3s_3=9$$
$$s_1=x_1+x_2+x_3$$ $$s_2=x_1 x_2 + x_1 x_3 + x_2 x_3$$ $$s_3=x_1 x_2 x_3$$ $$x_{1,2,3} \geq 0$$ Prove that: $s_2^4 \leq 72 s_3^2$
EDIT: As Michael Rozenberg said, this exercise is wrong. For $x_1=x_2=\frac{3}{2}$ and $x_3=0$ it is $(\frac{9}{4})^4\leq0$ which is wrong.
So, I think it should be: Prove that $s_2^4 \geq 72 s_3^2$
Can you help me, please? Thank you! I don't know how to start. Please help me or give maybe a hint. Also, please recommend or edit with a good title. Thank you!
It's wrong! Try $x_1=x_2=\frac{3}{2}$ and $x_3=0$.