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While solving one inequality, I arrived at a much simpler, but still nontrivial inequality $$(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$$ where $a,b,c$ are positive real numbers.

It apparently holds, but I can't seem to find a proof. The problem is it's not symmetric and applying inequalities $(a+b+c)^2\geq3(ab+bc+ca)$ or $a^3+b^3+c^3\geq ab^2+bc^2+ca^2$ won't work.

Any ideas?

user2345215
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  • Can you share you initial ineqaulity? Any additional infomration may help other users. – Stefan4024 Mar 12 '14 at 17:10
  • Are you sure your inequality is true? This comment is not a solution, but just a few rmks that might not lead to a solution but might rule out some approaches.

    We see $(ab+bc+ca)(ab^2 + bc^2 + ca^2) = abc(ab + bc + ca)(b/c + c/a + a/b).$ The function $x \mapsto x^{-1}$ is convex on $(0,\infty)$ and so Jensen implies $(a+b+c)^2 \leq (ab+bc+ca)(b/c+c/a+a/b).$ On the other hand, AM-GM tells us that $abc \leq (a+b+c)^3/27.$ Both these inequalities seem to be sharp. So either the inequality you posed is very sharp and a simple approach like this isn't going to work or the ineq. isn't ok.

     – Raghav Mar 12 '14 at 17:16
  • Here's an idea. (I don't know if it helps.) Let $f(a,b,c)=(a+b+c)^5-27(ab+bc+ca)(ab^2+bc^2+ca^2)$. You need to show that the minimum value of $f$ for $a,b,c\in\mathbb{R}^+$ is nonnegative. Suppose the minimum occurs at $(a_0,b_0,c_0)$. Note that $f$ is homogeneous: $f(ka,kb,kc)=f(a,b,c)$, so WLOG assume $a_0=1$ and minimize $f(b,c)=(1+b+c)^5-27(b+bc+c)(b^2+bc^2+c)$. – Steve Kass Mar 12 '14 at 17:19
  • @Stefan4024: http://math.stackexchange.com/questions/707212/how-to-prove-four-variable-inequality-involving-sums-of-cube-roots – user2345215 Mar 12 '14 at 17:21
  • @Raghav: But you are right, it seems to be rather tight. I can't quite see the Jensen, but it's a simple Cauchy $(ab+bc+ca)(\frac ab+\frac bc+\frac ca)\ge(a+b+c)^2$. So you are right it's a product of 2 competing inequalities, but it's still one way in the end, I have verified that with Mathematica both formally and numerically. – user2345215 Mar 12 '14 at 18:38
  • I don't have a solution, but we have the inequality $(a+b+c)^3 \geq \frac{27}{4} (ab^2 + bc^2 + ca^2 + abc)$, while you wish to prove that $(a+b+c)^3 \geq 9 (ab^2 + bc^2 + ca^2)$. Perhaps the two are related? – MT_ Mar 13 '14 at 03:14
  • @user92774 Your last inequality is not true, consider $a=b=1,c=0$. – user2345215 Mar 13 '14 at 08:59
  • @user2345215 true, but if you let $a+b+c = 1$ and $a, b, c \in \mathbb{R}^+$ the maximum is reached at $\frac{1}{9}$ – MT_ Mar 13 '14 at 12:13

1 Answers1

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Without loss of generality, let $a+b+c=3$. Then we only have to prove $$(a^2b+b^2c+c^2a)(ab+bc+ac)\le 9$$ Assume $b=\text{mid}(a,b,c)$, then we have $$c(b-a)(b-c)\le 0\Longleftrightarrow a^2b+b^2c+c^2a\le b(a^2+ac+c^2)$$ So $$(a^2b+b^2c+c^2a)(ab+bc+ac)\le b(a^2+ac+c^2)(ab+bc+ac)$$ Using AM-GM inequality, we get $\begin{align*} b(a^2+ac+b^2)(ab+bc+ca)&\le\,b\cdot\dfrac{(a^2+ac+c^2+ab+bc+ac)^2}{4}= b\cdot\dfrac{(a+c)^2(a+b+c)^2}{4}\\ &=\dfrac{9}{8}\cdot 2b\cdot(a+c)\cdot(a+c)\\ &\le\dfrac{9}{8}\cdot\dfrac{(2b+a+c+a+c)^3}{27}\\ &=\,9 \end{align*}$ Done!

user2345215
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math110
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  • Very nice! There's just a small typo in the last inequality, it should be a 3rd power, but otherwise it's perfect! – user2345215 Mar 13 '14 at 08:52