<s>[INST] Let $a,b,c$ be positive real numbers then prove the inequality $\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{81}$ For positive real numbers $a, b, c$ prove that
$$\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{81}$$
I subjected $a+1$, $1+\frac{b}{a}$, $1+\frac{c}{b}$ and $1+\frac{16}{c}$ to AM-GM inequality and obtained the following equations
$$a+1≥2\sqrt{a}$$
$$1+\frac{b}{a}≥2\sqrt{\frac{b}{a}}$$
$$1+\frac{c}{b}≥2\sqrt{\frac{c}{b}}$$
$$1+\frac{16}{c}≥2\sqrt{\frac{16}{c}}$$
Multiplying, we get,
$$(a+1)(1+\frac{b}{a})(1+\frac{c}{b})(1+\frac{16}{c})≥16\sqrt{16}$$
or $$\frac{(a+1)(a+b)(b+c)(c+16)}{abc}≥64$$
i.e.$$\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{64}$$
I am not arriving at the desired result and have realised that the question should be solved in another way. Please let me know how to solve this and also mention where I missed out.
 [/INST] Take $a=2x$, $b=4y$ and $c=8z$ and use AM-GM.
Indeed, we need to prove that
$$(1+a)(a+b)(b+c)(c+16)\geq81abc$$ or
$$(1+2x)(2x+4y)(4y+8z)(8z+16)\geq81\cdot64xyz$$ or
$$(1+2x)(x+2y)(y+2z)(z+2)\geq81xyz,$$
which is AM-GM:
$$(1+2x)(x+2y)(y+2z)(z+2)\geq3\sqrt[3]{x^2}\cdot3\sqrt[3]{xy^2}\cdot3\sqrt[3]{yz^2}\cdot3\sqrt[3]{z}=81xyz.$$
Done!
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