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Let $x$, $y$ and $z$ be positive numbers. Prove that: $$\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}.$$

This problem is similar to very many contest problems, but I think it's hard enough.

I tried to use Holder, C-S, AM-GM and more, but without any success.

For example, by Holder $$\sum_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}\right)^2\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}},$$ but I did not find non-negatives $k$ and $m$, for which the inequality $$4\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3\geq45\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3$$ is true.

Thank you!

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    $\begingroup$ Why don't you want to reduce the number of variables? For example, by dividing the numerator and denominator by $x^2$, we get $2$ variable inequality by substitute $\frac yx=m, \frac zy=n$ I do not claim that this solves the inequality. But at least it looks simpler. $\endgroup$
    – User
    Commented Nov 12, 2022 at 20:17
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    $\begingroup$ @User By your idea we'll loose homogeneous. Thank you for your interest! $\endgroup$ Commented Nov 12, 2022 at 20:25
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    $\begingroup$ @MichaelRozenberg Perhaps it is good to contain text in the title since one may right click the title. $\endgroup$
    – River Li
    Commented Nov 13, 2022 at 6:23
  • $\begingroup$ @River Li I don't know how to do it. $\endgroup$ Commented Nov 13, 2022 at 7:48
  • $\begingroup$ @MichaelRozenberg For example, add "Prove that". $\endgroup$
    – River Li
    Commented Nov 13, 2022 at 7:55

1 Answer 1

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(WARNING: this proof is very bad) First, define two sequences $a_i$ and $b_i$ where $a_1=4x^2+y^2,a_2=4y^2+z^2,a_3=4z^2+x^2$ and $b_1=\frac{1}{3x^2+yz},b_2=\frac{1}{3y^2+xz},b_3=\frac{1}{3z^2+xy}$. Since we have two three term sequences, we may now use Holder's inequality: $$\sqrt{4x^2+y^2+4y^2+z^2+4z^2+x^2}=\sqrt{5x^2+5y^2+5z^2}$$ and $$\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}$$ Thus, by Holder's inequality: $$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right) \ge \frac{3\sqrt{5}}{2}$$ Since equality holds when the two sequences are proportional, we may set $x=y=z$ and get: $$\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}$$ which is the minimum value of this inequality, we may then say that since: $$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right)=\sqrt{\frac{4x^2+y^2}{3x^2+yz}} + \sqrt{\frac{4y^2+z^2}{3y^2+xz}} + \sqrt{\frac{4z^2+x^2}{3z^2+xy}}$$ This holds to be true.

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