Let $a,b,c\ge0$, prove that: $$(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\ge16abc$$
I tried AM-GM as below:
$$a^2+1\ge2a; a^2+b^2\ge2ab; a^2+b^2+c^2\ge ab+bc+ca$$ The rest is proving the inequality which is not true for all non-negative real numbers: $a(2ab+1)(ab+bc+ca+1)\ge8abc$
By the way, equality do not hold for a=b=c, so I guess my approach seems lost. Is there any better idea to help me deal with problem? Thank you!
Best Answer
If you write it as a quadratic in $c$ and find its discriminant, then it suffices to prove: $$(a^2+1)^2(a^2+b^2+1)^3\geq 64a^2b^2.$$ This can be done with AM-GM since: $$(a^2+1)^2 = \left(a^2+\frac 13+\frac 13+\frac 13\right)^2\geq (4\cdot3^{- 3/4}a^{ 1/2})^2=\dfrac{16}{3\sqrt{3}}a$$ and so $$(a^2+b^2+1)^3 = \left(b^2+\dfrac{a^2+1}{2}+\dfrac{a^2+1}{2}\right)^3\geq \dfrac{27}{4}b^2(a^2+1)^2\geq12\sqrt{3}b^2a$$ and when you multiply them together, we obtain the desired. The equality is reached at: $$b = \sqrt{2}a = \dfrac{\sqrt{2}}{\sqrt{3}}$$ and from this you can deduce when is your original inequality is attained.
EDIT:
If one insists on using purely AM-GM, then: $$c^2+a^2+b^2+1\geq 2c\sqrt{a^2+b^2+1}$$ and therefore: $$(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\geq 2c(a^2+1)\sqrt{(a^2+b^2+1)^3}\geq$$ $$\geq 2c(a^2+1)\dfrac{3\sqrt{3}}{2}b(a^2+1)=3\sqrt{3}bc(a^2+1)^2\geq 3\sqrt{3}bc\cdot\dfrac{16}{3\sqrt{3}}a=16abc.$$