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

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!
