Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$

Question

Let $a,b,c\ge 0$ be such that $a^2+b^2+c^2=3$. Show that $$(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$$

I want to consider the function $$f(x)=\ln{(x^{3/2}+x^{1/2}+1)}$$ Maybe it isn't the case $f''(x)\le 0$, so I can't use Jensen's inequality.

Answer 1

Your idea works, but in another writing: $$3\ln3-\prod_{cyc}(a^3+a+1)=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)\right)=$$ $$=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)+\frac{2}{3}(a^2-1)\right)\geq0$$ because easy to see that $f(x)\geq0$ for any $x\geq0$, where $$f(x)=\ln3-\ln(x^3+x+1)+\frac{2}{3}(x^2-1).$$ Indeed, $$f'(x)=\frac{4(x-1)(2x+3)(2x^2-x+1)}{3(x^3+x+1)},$$ which gives $x_{min}=1$, $f(1)=0$ and we are done!

Answer 2

Let $f,g$ be given by $$ \left\lbrace \begin{align*} f(a,b,c)&=(a^3+a+1)(b^3+b+1)(c^3+c+1)\\[4pt] g(a,b,c)&=a^2+b^2+c^2\\[4pt] \end{align*} \right. $$ Using the method of Lagrange Multipliers, if a point $(a,b,c)$ satisfying the constraint $g(a,b,c)=3$ is such that $f$ has a local extremum at $(a,b,c)$, then for some $\lambda\in\mathbb{R}$ we must have $$ \left\lbrace \begin{align*} (3a^2+1)(b^3+b+1)(c^3+c+1)&=2\lambda a&&(\text{eq}1)\\[4pt] (a^3+a+1)(3b^2+1)(c^3+c+1)&=2\lambda b&&(\text{eq}2)\\[4pt] (a^3+a+1)(b^3+b+1)(3c^2+1)&=2\lambda c&&(\text{eq}3)\\[4pt] a^2+b^2+c^2&=3&&(\text{eq}4)\\[4pt] \end{align*} \right. $$ Given $a,b,c\ge 0$, each of equations $1,2,3$ has positive $\text{LHS}$, so must also have positive $\text{RHS}$, hence $a,b,c,\lambda > 0$.

Without loss of generality, assume $a=\max(a,b,c)$.

Then from $(\text{eq}4)$, it follows that $1\le a < \sqrt{3}$.

Suppose $a\ne b$.

Dividing $(\text{eq}1)$ by $(\text{eq}2)$, we get \begin{align*} & \frac {(3a^2+1)(b^3+b+1)} {(3b^2+1)(a^3+a+1)} = \frac{a}{b} \\[4pt] \implies\; & a(a^3+a+1)(3b^2+1)=b(b^3+b+1)(3a^2+1) \\[4pt] \implies\; & a(a^3+a+1)(3b^2+1)-b(b^3+b+1)(3a^2+1)=0 \\[4pt] \implies\; & (a-b)(3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2-3ab+a+b+1)=0 \\[4pt] \implies\; & 3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2-3ab+a+b+1=0 \\[4pt] \implies\; & 3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2+a+b=3ab-1 \\[4pt] \implies\; & (a+b)(3a^2b^2+a^2+b^2+1)=3ab-1 \\[4pt] \implies\; & (b+1)(4b^2+2)< 6b-1 \quad\text{[replaced $a$ by $1$ on the left and by $2$ on the right]} \\[4pt] \implies\; & 4b^3+4b^2-4b+3 < 0 \\[4pt] \implies\; & (4b^2-4b+1)+(4b^3+2) < 0 \\[4pt] \implies\; & (2b-1)^2+(4b^3+2) < 0 \\[4pt] \end{align*} contradiction.$\;$Hence $a=b$.

Analogously we get $a=c$.

Then $a=b=c$, hence from $(\text{eq}4)$, we get $a=b=c=1$.

It follows that for $a,b,c\ge 0$, the maximum value of $f(a,b,c)$ subject to the constraint $g(a,b,c)=3$ is $f(1,1,1)=27$.