Inductively. If a, b, c ≥ 1, prove that 4(abc + 1) ≥ (1 + a) (1 + b) (1 + c)

Question

Any idea to prove this problem by induction?. If $$a, b, c ≥ 1,$$ prove that $$4(abc + 1) ≥ (1 + a) (1 + b) (1 + c)$$(Suggestion- Rencia: Prove, more generally, that $${{{2^n}^-}^1} (a_1a_2\cdots a_n+1) ≥ (1+a_1)(1+a_2)\cdots(1+a_n).)$$

Answer 1

HINT

We have that

• base case $n=1 \implies (a_1+1)\ge (a_1+1)$
• inducion step, we assume true ${2^{n-1}} (a_1a_2...a_n+1) \ge (1+a_1)(1+a_2)...(1+a_n)$ and we need to prove that ${2^{n}} (a_1a_2...a_{n+1}+1) \ge (1+a_1)(1+a_2)...(1+a_{n+1})$, then consider

$${2^{n}} (a_1a_2...a_{n+1}+1)=2\cdot2^{n-1}a_1a_2...a_{n+1}+2^n\stackrel{Ind. Hyp.}\ge 2\left((1+a_1)(1+a_2)...(1+a_n)-2^{n-1}\right)a_{n+1}+2^n=(1+a_1)(1+a_2)...(1+a_n)\cdot 2a_{n+1}-2^na_{n+1}+2^n\stackrel{?}\ge (1+a_1)(1+a_2)...(1+a_{n+1})$$

then to complete the proof we need to check the following

$$(1+a_1)(1+a_2)...(1+a_n)\cdot 2a_{n+1}-2^na_{n+1}+2^n\stackrel{?}\ge (1+a_1)(1+a_2)...(1+a_{n+1})$$