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Let $a$, $b$ and $c$ be non-negative numbers such that $a^4+b^4+c^4=3$. Prove that: $$\frac{1}{3-2ab}+\frac{1}{3-2ac}+\frac{1}{3-2bc}\leq3$$ I tried SOS, C-S and more, but without success.

By brute force methods with computer we can get that this inequality is true, but I am looking for an human proof, which we can release during a competition.

Thank you!

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  • $\begingroup$ Oh, I see the problem now. $\endgroup$ – Sangchul Lee Dec 25 '16 at 4:28
  • $\begingroup$ @Sangchul Lee Thank you for your interest! $\endgroup$ – Michael Rozenberg Dec 25 '16 at 4:31
  • $\begingroup$ I have a partial proof of this do you want to see that ? Moreover it's short and clear .Perhaps this could inspire you... :) $\endgroup$ – user448747 Aug 14 '17 at 16:13
  • $\begingroup$ @FatsWallers I think it's better to give a full proof. $\endgroup$ – Michael Rozenberg Aug 14 '17 at 16:33
  • $\begingroup$ Before I post my proof are you agree with this fact : the maximum of $ab$ with the two conditions $a^4+b^4+c^4=3$ and $a\geq b \geq c $ is $$\sqrt{\frac{3}{2}}$$ ? $\endgroup$ – user448747 Aug 14 '17 at 20:20
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First of all we prove this inequality with $a^4+b^4+c^4=3$:

$$\frac{2}{3-a(b+c)}+\frac{1}{3-2bc}\leq 3$$

If we put this with $a\geq b \geq c$ :

$a=A$

$b=AB$

$c=AC$

Your inequality is equivalent to this :

$$\frac{2\sqrt{1+B^4+C^4}}{\sqrt{3}(\sqrt{3}\sqrt{1+B^4+C^4}-B-C)}+\frac{\sqrt{1+B^4+C^4}}{\sqrt{3}(\sqrt{3}\sqrt{1+B^4+C^4}-2BC)}$$

Or:

$$\frac{2(B+C)}{\sqrt{3}(\sqrt{3}\sqrt{1+B^4+C^4}-B-C)}+\frac{BC}{\sqrt{3}(\sqrt{3}\sqrt{1+B^4+C^4}-2BC)}+1$$

Now the idea is to use an identity :

$B^4+C^4=((B+C)^2-2BC)^2-2B^2C^2$

The inequality becomes :

$\frac{2(B+C)}{3(\sqrt{3}\sqrt{1+((B+C)^2-2BC)^2-2B^2C^2}-B-C)}+\frac{BC}{3(\sqrt{3}\sqrt{1+((B+C)^2-2BC)^2-2B^2C^2}-2BC)}+1$

Notice that we have just in the inequality elements of the form $BC$ or $B+C$

So we put :

$B+C=\lambda=constant$

And now remark that $bc$ have his maximum when $B=\frac{\lambda}{2}$ and $C=\frac{\lambda}{2}$

It's the same thing in the general inequality (we can replace $BC$ by $BC=\frac{\lambda^2}{4}$) so we have this inequality with one variable :

$$\frac{2(\lambda)}{3(\sqrt{3}\sqrt{1+((\lambda)^2-2\frac{\lambda^2}{4})^2-2(\frac{\lambda^2}{4})^2}-(\lambda))}+\frac{\frac{\lambda^2}{4}}{3(\sqrt{3}\sqrt{1+((\lambda)^2-2\frac{\lambda^2}{4})^2-2(\frac{\lambda^2}{4})^2}-2\frac{\lambda^2}{4})}+1$$

Wich is equivalent to :

$$f(\lambda)=\frac{2(\lambda)}{3(\sqrt{3}\sqrt{1+\frac{\lambda^4}{8})}-(\lambda))}+\frac{\frac{\lambda^2}{4}}{3(\sqrt{3}\sqrt{1+\frac{\lambda^4}{8}}-2\frac{\lambda^2}{4})}+1$$

Now it's easy to remark that the function have a maximum for $\lambda=2$ wich is $3$

So we have prove the following inequality with some conditions :

$$\frac{2}{3-a(b+c)}+\frac{1}{3-2bc}\leq 3$$

Now the idea is to combine this with the theorem of logarithmic majorization for three variables :

Theorem :

Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $e\geq d \geq f $ under the three following conditions :

$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :

$$a+b+c\geq d+e+f$$

The inequality to prove is :

$$\sum_{cyc}\frac{1}{2-3ab}\leq \frac{2}{3-e(f+d)}+\frac{1}{2-3df}\leq 3 $$

With $a^4+b^4+c^4=e^4+f^4+d^4=3$

And $d=1$,$f=bc$,$e=(2-(bc)^4)^{\frac{1}{4}}$

Now its easy to see that we have :

$\frac{1}{2-3ab}\geq \frac{1}{2-3ac}\geq \frac{1}{2-3bc}$

And

$\frac{1}{3-e(f+d)}\geq \frac{1}{2-3df}$

But as we have said in a comment the maximum of $ab$ is $\sqrt{1.5}$ and $0\leq bc \leq 1$

And :

$e(f+d)=(2-(bc)^4)^{\frac{1}{4}}(1+bc)\geq \sqrt{1.5}$ for $bc\geq0.5$

So we have :

$\frac{1}{3-e(f+d)}\geq \frac{1}{2-3\sqrt{1.5}}\geq\frac{1}{2-3ab}\geq \frac{1}{2-3ac}$

So you just have to apply the theorem to have :

$$\sum_{cyc}\frac{1}{2-3ab}\leq \frac{2}{3-e(f+d)}+\frac{1}{2-3df}\leq 3 $$

For the case $bc\leq 0.5$ you just have to change the parameters like this :

$d=\frac{1}{2}$,$f=2bc$,$e=(2-\frac{1}{16}-(16bc)^4)^{\frac{1}{4}}$

So we have :

$e(f+d)=(2-\frac{1}{16}-(16bc)^4)^{\frac{1}{4}}(1+bc)\geq \sqrt{1.5}$ for $bc\leq0.5$

And the reasoning is the same.

Done!

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  • $\begingroup$ Now it would right... :) $\endgroup$ – user448747 Aug 15 '17 at 20:30

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