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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.

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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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