For $a,b,c>$ and $\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\le 1$. Prove that $$ab^2c^3d^5\le \frac{1}{10^{11}}$$
$\frac{a}{{1 + a}} + \frac{{2b}}{{1 + b}} + \frac{{3c}}{{1 + c}} \le 1 \leftrightarrow \frac{1}{{1 + a}} + \frac{2}{{1 + b}} + \frac{3}{{1 + c}} \ge 5$
$= > \frac{1}{{a + 1}} \ge 2 - \frac{2}{{1 + b}} + 3 - \frac{3}{{1 + b}} = \frac{{2b}}{{1 + b}} + \frac{{3c}}{{1 + c}}$
$\ge 5\sqrt[5]{\frac{b^2}{(1+b)^2}*\frac{c^3}{(1 + c)^3}}$
$\frac{1}{1 + b} \ge \frac{a}{1 + a} + \frac{b}{1 + b} + \frac{3c}{1 + c} \ge 5\sqrt[5]{\frac{a}{1 + a}*\frac{b}{1 + b}*\frac{c^3}{(1+c)^3}}$
$\frac{1}{1 + c} \ge \frac{a}{1 + a} + \frac{2b}{1 + b} + \frac{2c}{1 + c} \ge 5\sqrt[5]{\frac{a}{1 + a}.\frac{b^2}{(1+b)^2}.\frac{c^2}{(1+c)^2}}$
$=> \frac{1}{1 + a}*{(\frac{1}{1 + b} )^2}*(\frac{1}{1 + c})^3\ge 5^6\frac{a(b^2)(c^3)}{({1 + a} )}{({1 + b})^2}{{( {1 + c} )}^3} $
Seem wrong, fix for me
I think it means the following inequality.
Let $a$, $b$, $c$ and $d$ be positive numbers such that $$\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\leq1.$$ Prove that $$ab^2c^3d^5\leq\frac{1}{10^{11}}.$$ I think your idea works here very well.
From the condition by AM-GM we obtain: $$\frac{1}{1+a}\geq\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\geq10\sqrt[10]{\frac{b^2c^3d^5}{(1+b)^2(1+c)^3(1+d)^5}};$$ $$\frac{1}{1+b}\geq\frac{a}{1+a}+\frac{b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\geq10\sqrt[10]{\frac{abc^3d^5}{(1+a)(1+b)(1+c)^3(1+d)^5}};$$ $$\frac{1}{1+c}\geq\frac{a}{1+a}+\frac{2b}{b+1}+\frac{2c}{c+1}+\frac{5d}{d+1}\geq10\sqrt[10]{\frac{ab^2c^2d^5}{(1+a)(1+b)^2(1+c)^2(1+d)^5}}$$ and $$\frac{1}{1+d}\geq\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{4d}{d+1}\geq10\sqrt[10]{\frac{ab^2c^3d^4}{(1+a)(1+b)^2(1+c)^3(1+d)^4}}.$$ Thus, $$\frac{1}{(1+a)(1+b)^2(1+c)^3(1+d)^5}\geq10^{1+2+3+5}\sqrt[10]{\frac{a^{2+3+5}b^{2+2+6+10}c^{3+6+6+15}d^{5+10+15+20}}{(1+a)^{10}(1+b)^{20}(1+c)^{30}(1+d)^{50}}}$$ or $$ab^2c^3d^5\leq\frac{1}{10^{11}}$$ and we are done!