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 :-)

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    $\begingroup$ I don't know how to solve it, but maybe this can help yout: $$ a^2 + b^2 + c^2 = 3 => abc \le 1 $$ $\endgroup$ – user59169 Jan 22 '13 at 0:19
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    $\begingroup$ @daniel: $a=b=c=1$ is a solution. $\endgroup$ – Eckhard Jan 22 '13 at 0:26
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    $\begingroup$ 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. $\endgroup$ – Eckhard Jan 22 '13 at 1:09
  • $\begingroup$ 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. $\endgroup$ – JavaMan Jan 22 '13 at 5:57
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    $\begingroup$ 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) $\endgroup$ – user27126 Jan 22 '13 at 6:23

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 $u=v=w\Rightarrow a=b=c$

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

  • $\begingroup$ How do you know that the max occurs when $a=b=c$? $\endgroup$ – Eckhard Jan 22 '13 at 7:47
  • $\begingroup$ @Eckhard When the max occurs, we have $u=v=w\Rightarrow a=b=c$ $\endgroup$ – Shane Chern Jan 22 '13 at 7:54
  • $\begingroup$ How do you know that the max occurs when $u=v=w$? $\endgroup$ – Eckhard Jan 22 '13 at 8:01
  • $\begingroup$ @Eckhard Er... that is what Jensen's inequality indicates. $\endgroup$ – Shane Chern Jan 22 '13 at 8:51
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    $\begingroup$ @Eckhard I've checked it in Hardy's Inequalities, 2nd ed, and that's exactly what the equality cases are as no interval are linear (see in §3.8). $\endgroup$ – Shane Chern Jan 22 '13 at 10:24

Final solution :


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

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

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


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 $27$.

  • $\begingroup$ How is this different from Lagrange multiplier? This is just phrased in a tone-down way I suppose. $\endgroup$ – user27126 Jan 25 '13 at 18:13
  • $\begingroup$ 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. $\endgroup$ – robjohn Jan 25 '13 at 18:31

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.

  • $\begingroup$ 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. $\endgroup$ – robjohn Jan 26 '13 at 20:54
  • $\begingroup$ Well, what I had in mind was differentiation, which is quite easy in this particular case $\endgroup$ – leshik Jan 28 '13 at 16:09
  • $\begingroup$ 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. $\endgroup$ – robjohn Jan 28 '13 at 16:28

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.

  • $\begingroup$ How is this different from the answer with 7 votes already? $\endgroup$ – user27126 Jan 28 '13 at 8:49
  • $\begingroup$ It's not. I didn't pay enough attention. $\endgroup$ – ivan Jan 28 '13 at 11:20

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:

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

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    $\begingroup$ 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! $\endgroup$ – user27126 Feb 8 '13 at 18:09

It is enough to show that for $t\geqslant0$, $$f(t) = \log(3)-\log(a^3+a+1)+\tfrac23(t^2-1) \geqslant 0$$ 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 $f(t)\geqslant f(1)=0$.


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.

  • $\begingroup$ I think you mean it's Schur concave. Why is it true though? $\endgroup$ – user27126 Jan 22 '13 at 5:23
  • $\begingroup$ It is symmetric and monotonically increasing in each argument. $\endgroup$ – dineshdileep Jan 22 '13 at 5:25
  • $\begingroup$ sorry, i meant it is schur-concave. Thanks for the help $\endgroup$ – dineshdileep Jan 22 '13 at 5:26
  • $\begingroup$ How does monotonically increasing imply concavity? When one vector majorizes another, it's possible that some components are larger, while some components are smaller. $\endgroup$ – user27126 Jan 22 '13 at 5:27
  • $\begingroup$ 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. !! $\endgroup$ – dineshdileep Jan 22 '13 at 5:34

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