parmenides51 wrote: Let $a,b,c>0$ and $abc=1$ . Prove that $$ \sqrt{2(1+a^2)(1+b^2)(1+c^2)}\ge 1+a+b+c.$$ Solution Prove that for any real numbers $a,b,c:$$$ \sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge a+b+c+ab+bc+ca-1-abc \ge 1+a+b+c. $$

$$\sqrt{2(1 + a^2)(1 + b^2)(1 + c^2)} = \sqrt{(a^2 + 1 + 1 + a^2)((bc)^2 + b^2 + c^2 + 1)} \ge abc + b + c + a = 1 + a + b + c$$