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Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that: $$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$

I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and $$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.

Thanks :-)

Iuli
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    @daniel: $a=b=c=1$ is a solution. – Eckhard Jan 22 '13 at 00:26
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    I don't know how to solve it, but maybe this can help yout: $$ a^2 + b^2 + c^2 = 3 => abc \le 1 $$ – user59169 Jan 22 '13 at 00:19
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    One route to a solution would be to use the method of Lagrange multipliers to find the max of the function $(a^3+a+1)(b^3+b+1)(c^3+c+1)$ and to observe that of the $56$ (or so) possible candidates only $(1,1,1)$ has three positive coordinates. – Eckhard Jan 22 '13 at 01:09
  • It may be worth noting that $x^3 + x + 1 \leq 3$ for all $x \leq 1$. If $a \geq b \geq c \geq 0$, then you need only worry about when $\sqrt{3} \geq a > 1$ (and hence $c < 1$). I suppose this is the hardest one needs to handle here. – JavaMan Jan 22 '13 at 05:57
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    This inequality can be solved by noting that $x \to \ln(x^{3/2}+x^{1/2}+1)$ is concave for $x > 0$ and Jensen's inequality. Since dineshdileep noticed something similar first, I won't post this as answer. (Unless he doesn't edit his answer) –  Jan 22 '13 at 06:23
  • solutions without derivatives ? – Iuli Jan 22 '13 at 06:48
  • @Iuli: It is not considered good form to make minor edits to your question simply to keep it on the front page. Note that after too many edits, your post will become Community Wiki, and you will stop getting reputation for it. – robjohn Jan 25 '13 at 13:11
  • @Iuli: You may think about the substitution $a=\sqrt{3}\cos \alpha$, $b=\sqrt{3}\cos \beta$ and $c=\sqrt{3}\cos \gamma$ where $\cos^2 \alpha+ \cos^2 \beta+ \cos^2 \gamma=1$. –  Nov 05 '14 at 13:42

8 Answers8

15

Let $u:=a^2, v:=b^2, w:=c^2$, we have $u+v+w=3$.
Consider the function $$f(x)=\ln (1+x^{\frac{1}{2}}+x^{\frac{3}{2}}),\ 0<x\leq 3$$ it's easy to compute that $f''(x)<0$.
by Jensen's inequality, we have $$\sum\ln (1+u^{\frac{1}{2}}+u^{\frac{3}{2}})\leq3f(\dfrac{\sum u}{3})=3\ln 3$$ that is $$\prod(a^3+a+1) \leq 27$$ When the max occurs, we have $u=v=w\Rightarrow a=b=c$
Q.E.D.

By the way, I draw a graph of $f''(x)$ on $(0,3]$ by mathematica to show it more directly...
enter image description here

Shane
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Final solution :

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=527572&p=2997801#p2997801

$$ (a-1)^4\geq0\Rightarrow a^3+a+1\leq\frac{(a^2+1)(a^2+5)}{4} $$

$$ \prod(a^3+a+1)\leq\frac{1}{64}\prod(a^2+1)\prod(a^2+5)$$

$$ \leq\frac{1}{64}\frac{(a^2+b^2+c^2+3)^3}{27}\frac{(a^2+b^2+c^2+15)^3}{27}=27 $$

Iuli
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For what it's worth, this answer uses derivatives, but it does use a generally applicable method.

Since $a^2+b^2+c^2=3$, any variations, $(\delta a,\delta b,\delta c)$, of $(a,b,c)$ must satisfy $$ a\,\delta a+b\,\delta b+c\,\delta c=0\tag{1} $$ We are interested in finding the maximum of $$ \log(a^3+a+1)+\log(b^3+b+1)+\log(c^3+c+1)\tag{2} $$ At a critical point, the variations of $(2)$ must satisfy $$ \frac{3a^2+1}{a^3+a+1}\delta a+\frac{3b^2+1}{b^3+b+1}\delta b+\frac{3c^2+1}{c^3+c+1}\delta c=0\tag{3} $$ Standard linearity arguments say that if $(3)$ is true for all $(\delta a,\delta b,\delta c)$ that satisfy $(1)$, we have $$ \left(\frac{3a^2+1}{a^3+a+1},\frac{3b^2+1}{b^3+b+1},\frac{3c^2+1}{c^3+c+1}\right)=k(a,b,c)\tag{4} $$ That is, $$ \frac{3a^2+1}{a^4+a^2+a}=\frac{3b^2+1}{b^4+b^2+b}=\frac{3c^2+1}{c^4+c^2+c}\tag{5} $$ Note that $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{3x^2+1}{x^4+x^2+x} =-\frac{6 x^5+4 x^3-3 x^2+2 x+1}{(x^4+x^2+x)^2}\tag{6} $$ If $x\ge1$, then $6x^5+4x^3\ge3x^2$ and if $0\le x\le1$, then $2x+1\ge3x^2$. Therefore, for all $x\ge0$, $(6)$ is negative. That is, $$ \frac{3x^2+1}{x^4+x^2+x}\tag{7} $$ is monotonic decreasing which, when combined with $(5)$, says that $$ a=b=c\tag{8} $$ $(8)$ says that $$ (a^3+a+1)(b^3+b+1)(c^3+c+1)=27\tag{9} $$ Condition $(8)$ assumes that (a,b,c) is not on the boundary, that is none are $0$. Suppose that $c=0$, then the same argument yields that $a=b=\frac12\sqrt6$ and therefore $$ (a^3+a+1)(b^3+b+1)(c^3+c+1)=\frac{83}{8}+\frac52\sqrt6\tag{10} $$ Suppose that $b=c=0$, then $a=\sqrt3$ and therefore $$ (a^3+a+1)(b^3+b+1)(c^3+c+1)=1+4\sqrt3\tag{11} $$ Comparing $(9)$, $(10)$, and $(11)$, the maximum is $27$.

robjohn
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  • How is this different from Lagrange multiplier? This is just phrased in a tone-down way I suppose. –  Jan 25 '13 at 18:13
  • This is just an unrolled version of a Lagrange multiplier, as you say. I was pushing things to use derivatives, I didn't want to assume knowledge of Lagrange multipliers, too. – robjohn Jan 25 '13 at 18:31
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General comment: as soon as one tries to use AM-GM for $x=a^3+a+1,$ $y=b^3+b+1$ and $z=c^3+c+1$ the inequality becomes wrong, since $$a^3+a+b^3+b+c^3+c+3\ge 2(a^2+b^2+c^2)+3=9.$$ Using Lagrange multiplayers, one can reduce this problem to the following system: $$(3a^2+1)(b^3+b+1)(c^3+c+1)=2\lambda a$$ $$(3b^2+1)(a^3+a+1)(c^3+c+1)=2\lambda b$$ $$(3c^2+1)(a^3+a+1)(b^3+b+1)=2\lambda c.$$

In other words, if $\lambda\ne 0,$ for the function $$f(x)=\frac{x(x^3+x+1)}{(3x^2+1)}$$ we have $f(a)=f(b)=f(c).$ It is easy to see, that $f$ is monotone for $x\ge 0$ so the only option is $a=b=c=1.$ The rest should be clear.

leshik
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  • What is the easy way to see that $f$ is monotone for $x\ge0$? I took the derivative and was able to reason that its sign did not change. I would be interested in an easier method. – robjohn Jan 26 '13 at 20:54
  • Well, what I had in mind was differentiation, which is quite easy in this particular case – leshik Jan 28 '13 at 16:09
  • So $\displaystyle f'(x)=\frac{6 x^5+4 x^3-3 x^2+2 x+1}{\left(3 x^2+1\right)^2}\ge0$ by the same argument I used. Too bad. – robjohn Jan 28 '13 at 16:28
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This will have derivatives. Substitute $a=\sqrt{x}, b=\sqrt{y}$ and $c=\sqrt{z}$. Then $x+y+z=3$. Consider the function $f(x)=\ln(x\sqrt{x}+\sqrt{x}+1)$. It is concave.

Hence Jensen yields: $f(x)+f(y)+f(z)<=3f(1)=3\ln{3}$

This is equivalent to what is asked.

ivan
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I have a new answer. We will apply the Cebysev's inequality.

$$(a^3+a+1)(b^3+b+1)\leq3(a^3b^3+ab+1)$$ So:

$$\left[(a^3+a+1)(b^3+b+1)(c^3+c+1)\right]^{2} \leq 27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1).$$

Now, we will prove that:

$$27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27^{2}.$$ Equivalent with:

$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27.$$

Now we use $AM \geq GM$ :

$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq \left(\frac{a^3b^3+b^3c^3+a^3c^3+ab+bc+ca+3}{3}\right)^{3}.$$

Now we will apply the following inequality to obtain the desired result:

$$a^3b^3+b^3c^3+a^3c^3 \leq \frac{(a^2+b^2+c^2)^{3}}{9}=3,$$ $$ab+bc+ca \leq a^2+b^2+c^2=3.$$

Iuli
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    What Chebyshev inequality is this? If it's the similarly ordered sequence one, then why is ${a^3,a,1}$ and ${b^3,b,1}$ similarly ordered? In fact, they are not! –  Feb 08 '13 at 18:09
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It is enough to show that for $t\geqslant0$, $$f(t) = \log(3)-\log\left(t^3+t+1\right)+\tfrac23\left(t^2-1\right) \geqslant 0$$ as the inequality in question is merely $f(a)+f(b)+f(c)\geqslant0$.

But $f’(t)=0$ only when $t=1$ where it has a minimum, so $f(t)\geqslant f(1)=0$.

Hanno
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Macavity
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This is wrong

If you are familiar with majorization, observe that the function is schur concave. Thus, its maximum occurs at a point where all variables are equal, and since that point exists in the constraint set (i.e. $a,b,c\geq 0$ and $a^2+b^2+c^2=3$), $a=b=c=1$ is the maxima. Thus, the inequality comes.

dineshdileep
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  • I think you mean it's Schur concave. Why is it true though? –  Jan 22 '13 at 05:23
  • It is symmetric and monotonically increasing in each argument. – dineshdileep Jan 22 '13 at 05:25
  • sorry, i meant it is schur-concave. Thanks for the help – dineshdileep Jan 22 '13 at 05:26
  • How does monotonically increasing imply concavity? When one vector majorizes another, it's possible that some components are larger, while some components are smaller. –  Jan 22 '13 at 05:27
  • my intuition was that the function is symmetric, and also it is convex in each variable. Therefore, it should be schur-convex, but now it seems schur-concave is what we needed. I don't know how to correct it, it all seemed to fit well. !! – dineshdileep Jan 22 '13 at 05:34
  • Product of convex functions may not be convex. In fact LHS is neither convex or concave on $[0,3]$. I'm not sure about Schur-convexity/concavity, but I don't think it's something you can handwave. –  Jan 22 '13 at 05:36
  • In fact, no, it can't be Schur-convex/concave, because $log(x^3+x+1)$ is neither convex or concave. This means that one can cook up counterexamples by picking points in the region where the function is convex then picking points where the function is concave. –  Jan 22 '13 at 05:39
  • let us continue this discussion in chat – dineshdileep Jan 22 '13 at 05:42
  • my bad! I should have done a change of variable. In fact $x \to \log(x^{3/2}+x^{1/2}+1)$ is concave on $x > 0$. I am not familiar with the terminology of Schur concavity, but usual Jensen's inequality would work. –  Jan 22 '13 at 06:04
  • so, what is the conclusion? – dineshdileep Jan 22 '13 at 08:25
  • See Shane Chern's answer. –  Jan 22 '13 at 08:28
  • So does it mean the function is schur-concave after all? – dineshdileep Jan 22 '13 at 08:35
  • $(a,b,c) \to (a^3+a+1)(b^3+b+1)(c^3+c+1)$, no. $(u,v,w) \to (u^{3/2}+u^{1/2}+1)(v^{3/2}+v^{1/2}+1)(w^{3/2}+w^{1/2}+1)$, yes. –  Jan 22 '13 at 08:36