5
$\begingroup$

$$2s_1^2-4s_2+3s_3=9$$

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

$\endgroup$
1
$\begingroup$

It's wrong! Try $x_1=x_2=\frac{3}{2}$ and $x_3=0$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.