$$s_1=x_1+x_2+x_3$$ $$s_2=x_1 x_2 + x_1 x_3 + x_2 x_3$$ $$s_3=x_1 x_2 x_3$$ $$x_{1,2,3} \geq 0$$ 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 $x_3=0$.

  • $\begingroup$ This is true! It is wrong! Thank you! So the exercise is wrong. Should it be: $s_2^4 \geq 72 s_3^2$ ? Thank you! $\endgroup$ – MM PP Aug 19 '16 at 10:18
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    $\begingroup$ Now it's true. The Contradiction method helps here. $\endgroup$ – Michael Rozenberg Aug 19 '16 at 11:17
  • $\begingroup$ I can't see how to prove that using contradiction. Can you please give me a hint? Thank you very, but very much! $\endgroup$ – MM PP Aug 19 '16 at 11:51
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    $\begingroup$ Let $s_2^2<6\sqrt2s_3$, $x_1=ka$, $x_2=kb$ and $x_3=kc$, where $k>0$ and $(ab+ac+bc)^2=6\sqrt2abc$. Thus, $k<1$ and we can use an homogenization. $\endgroup$ – Michael Rozenberg Aug 19 '16 at 14:27

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