Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$

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 :-)

Solution 1:

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...

Solution 2:

Final solution :

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

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

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

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

Solution 3:

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 $27$.

Solution 4:

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.