$a ^ 2 + b ^ 2 + c ^ 2 = 12$ - prove that $(a+1)(b+1)(c+1) \leq 27$

Question

I've got stuck on this problem :

Let $a, b, c \geq 0$ such that $a ^ 2 + b ^ 2 + c ^ 2 = 12$. Prove that $(a+1)(b+1)(c+1) \leq 27$.

I've thought that mean inequality ($HM \leq GM \leq AM \leq PM$ - referring to harmonic, geometric, arithmetic and square mean) may help. I've also considered $CBS$ (Cauchy - Buniakowski - Scwarz) inequality and Cebîşev inequality, but I didn't get it right.

I would be thankful for some hints.

Answer 1

$$\frac{a ^ 2 + b ^ 2}{2} + \frac{c ^ 2 + b ^ 2}{2} + \frac{a ^ 2 + c ^ 2}{2} = 12$$ $$\therefore 12 \ge ab + bc + ca$$ Also, $$\frac{a^2 + b^2 + c^2}{3} \ge (abc)^{\frac23}$$ $$\therefore 4^{\frac32} \ge abc$$ $$\therefore 8 \ge abc$$ And finally, RMS > A.M. gives $$(\frac{a^2 + b^2 + c^2}{3})^\frac12 \ge \frac{a + b + c}{3}$$ $$\therefore 6 \ge a+b+c$$ $$\therefore 12 + 8 + 6 + 1 \ge abc + ab + bc + ca + a + b + c + 1$$ $$\therefore 27 \ge (a+1)(b+1)(c+1)$$

Answer 2

You may use the GM-AM-QM inequality, i.e. to combine: $$ (a+1)(b+1)(c+1) \leq \left(\frac{a+b+c+3}{3}\right)^3 \tag{1}$$ with: $$ \frac{a+b+c}{3}\leq \sqrt{\frac{a^2+b^2+c^2}{3}}.\tag{2}$$

Answer 3

$24=a^2+b^2+c^2 + 12 \geq 4a+4b+4c$
$a^2+b^2+c^2+2a+2b+2c+3 \leq12+12+3=27$
$(a+1)^2+(b+1)^2+(c+1)^2 \leq27$
$3\ ^3\sqrt{(a+1)^2(b+1)^2(c+1)^2} \leq (a+1)^2+(b+1)^2+(c+1)^2 \leq27$
$ ^3\sqrt{(a+1)^2(b+1)^2(c+1)^2} \leq 9$
$(a+1)(b+1)(c+1) \leq 27$