Let $a, b,c$ be positive real numbers such that $a+b+c=1,$ prove $(a + b) ^ 2 (1 + 2c)(2a + 3c)(2b + 3c) \ge54abc$
I think it's to be solved by AM-GM inequality but I am not getting to the final solution
here's what I thought we have 54 on RHS so we need 27*2. So could be a possibility that 3 am gm of 3 terms and 1 am gm of 2 term but unable to make such a case. I tried applying am gm individual on each bracket and multiplying all the resultants but that doesn't seem to work
applying am-gm on $(a+b)^2$ we get $>=4ab$, next we write $1+2c$ as $a+b+3c$ and apply am-gm we get $>= 3$ cube root $3$ $abc$ now the next bracket we write $2a+3c$ as $a+a+3c$ and apply am gm we get$>=3$ cuberoot $3a^2c$. And finally $2b+3c$ as $b+b+3c$ we get $>= 3$ cuberoot $3b^2c.$ Multiplying all of them we get the expression$>=81 *4 ab(abc)$
Smoothing helps.
Indeed, we need to prove that: $$(a+b)^2(a+b+3c)(2a+3c)(2b+3c)\geq54abc(a+b+c)^2$$ or $$(a+b)^2(a+b+3c)(4ab+6(a+b)c+9c^2)\geq54abc(a+b+c)^2.$$ Now, let $a+b=constant$.
Thus, we need to prove a linear inequality of $ab$, which says that it's enough to prove our inequality for an extremal value of $ab$.
If $ab\rightarrow0^+$, so our inequality is obvious.
But by AM-GM $ab$ gets a maximal value for $a=b$.
Now, let $a=cx$.
Thus, it's enough to prove that: $$4x^2(2x+3)^3\geq54x^2(2x+1)^2$$ or $$(2x-3)^2(4x+3)\geq0,$$ which ends a proof.