One can show $4a^3+4 \geq (a+1)^3$, similarly for $b$ and $c$. The rest is AM-GM, equality occurs at $a=b=c=1$.

Show that for any positive real numbers $a,b$ the following inequality is true: $$2(a^2+b^2+2)\geq 2(a+1)(b+1)$$ dangerousliri wrote: Show that for any positive real numbers $a,b,c$ the following inequality is true: $$4(a^3+b^3+c^3+3)\geq 3(a+1)(b+1)(c+1)$$When does equality hold? Proof of Zhangyunhua:

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nice problem,very nice solutions!

teomihai wrote: nice problem,very nice solutions! Yeah.

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive real numbers .Prove that$$2^{n-1}(a^n_1+a^n_2+\cdots+a^n_n+n)\geq n(a_1+1)(a_2+1)\cdots (a_n+1).$$n=4

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If 4a^3+4>=(a+1)^3 we are done.We simplify this and we get a^3+1>=a^2+a which is equivalent to a^2-2a+1>=0 which is absolutely true

$4 \sum (a^3 + 1) \ge 4 \sum \frac{(a + 1)^3}{4} = \sum (a + 1)^3 \ge 3 \prod (a + 1)$, using Holder (or Chebyshev) and $AM-GM$, respectively