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Let $a,b,c\in \mathbb{R^+}$.

Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$

PS: I don't have ay ideas about this problem :(

Thanks

wythagoras
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Lê Tấn Khang
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  • As $\frac{9\sqrt[3]{abc}}{a+b+c}\le \frac{3(a+b+c)}{a+b+c}=3$, if possible, the statement is equivalent to $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq3$ – Mythomorphic May 30 '15 at 11:58
  • Take the limit as $a$ tends to 0 for fixed $b=c=1$. You get something less than 3. Since the function is continuous around about that point, the left hand side can't be $\ge3$ everywhere. – Alan U. Kennington May 30 '15 at 12:20
  • oh I see. Haven't considered it. – Mythomorphic May 30 '15 at 12:22
  • Try AM-GM concept. Should give the answer, although the expansion will be quite tedious. – Gummy bears May 30 '15 at 13:43

3 Answers3

1

By Holder $\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b^2+c^2}}\right)^3\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3\geq\left(\sum\limits_{cyc}(a^2+ab)\right)^4$. Thus, it remains to prove that $\left(\sum\limits_{cyc}(a^2+ab)\right)^4(a+b+c)^3\geq729abc\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3$, which is obvious.

Michael Rozenberg
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    Nice +1, but hardly obvious. – Macavity May 31 '15 at 07:39
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    I don't get it. How is $\left(\sum\limits_{cyc}(a^2+ab)\right)^4(a+b+c)^3\geq729abc\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3$ obvious ? – Ewan Delanoy May 31 '15 at 07:47
  • Dear Ewan Delanoy! – Michael Rozenberg May 31 '15 at 14:03
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    Dear Ewan Delanoy! Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Hence, the last inequality is equivalent to $f(v^2)\geq0$, where $f(v^2)=(3u^2-v^2)^4u^3+w^3(18u^2w^6+20v^2w^6+162u^5w^3-135u^3v^2w^3-81uv^4w^3-243u^4v^4+243u^2v^6)$. Easy to show that $f$ is a decreasing function. Hence, $f$ gets a minimal value for maximal value of $v^2$, which happens, when two variables are equal. Thus, it remains to check the last inequality for $b=c=1$, which gives something obvious. – Michael Rozenberg May 31 '15 at 14:11
1

$\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge 3 \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \\ \iff \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \ge 3\dfrac{\sqrt[3]{abc}}{a+b+c} $

now we prove a stronger one :

$\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge \dfrac{3\sqrt[3]{abc}}{a+b+c} \iff \dfrac{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}{(b^2+c^2)(a^2+c^2)(a^2+b^2)} \ge \dfrac{3^3abc}{(a+b+c)^3} \iff \dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$

there is a lemma:

$\dfrac{(a+b)(b+c)(c+a)}{8abc} \ge \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$

Michael Rozenberg give this lemma a nice proof:

$(c^{2}+ab)(a+b)=a(b^2+c^2)+b(a^2+c^2) \ge 2 \sqrt{ab(b^2+c^2)(a^2+c^2)}$

with AM-GM, it is easy to prove:

$\dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(a+b)(b+c)(c+a)}{8abc}$

chenbai
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0

This is just a hunch. First, the terms are all homogenous of the same degree. That's a good start. Put $a=b=c=1$. This gives exact equality. That's probably useful for something. Chances are, that's the maximum or the minimum of the difference between left and right hand side. Now vary $a$ by reducing it. Take the derivative. You can assume initially that $b=c=1$. Reduce $a$ from 1 to 0. Then the inequality gets stronger by the time you get to $a=0$. My hunch is that it's probably monotonic all the way from $a=1$ down to $a=0$. You can do the same sort of thing with the pair $(b,c)$, showing convexity or concavity. Then the result should magically appear! That's just a hunch.

PS. This is what I think the graph looks like. cube root graph

That might give a few clues!

Alan U. Kennington
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