Question Number 160375 by HongKing last updated on 28/Nov/21
Commented by ghimisi last updated on 28/Nov/21
$$\frac{\mathrm{1}}{\mathrm{2}{b}+\mathrm{1}}\:{or}\:\frac{\mathrm{1}}{\mathrm{4}{b}+\mathrm{1}} \\ $$$$ \\ $$
Commented by HongKing last updated on 28/Nov/21
$$\mathrm{Sorry}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\frac{\mathrm{1}}{\mathrm{4b}\:+\:\mathrm{1}} \\ $$
Answered by ghimisi last updated on 28/Nov/21
$$\forall{t}\in{R},\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)\geqslant\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{3}{t}^{\mathrm{3}} +\mathrm{3}{t} \\ $$$${t}=\frac{{x}^{{a}} }{{x}^{{b}} }\Rightarrow{x}^{\mathrm{4}{a}} +\mathrm{4}{x}^{\mathrm{2}{a}+\mathrm{2}{b}} +{x}^{\mathrm{4}{b}} \geqslant\mathrm{3}{x}^{\mathrm{3}{a}+{b}} +\mathrm{3}{x}^{{a}+\mathrm{3}{b}} \\ $$$$\Rightarrow\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{4}{a}} {dx}+\mathrm{4}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{2}{a}+\mathrm{2}{b}} {dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{4}{b}} {dx}\geqslant\mathrm{3}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{3}{a}+{b}} {dx}+\mathrm{3}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{{a}+\mathrm{3}{b}} {dx} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}{a}+\mathrm{1}}+\frac{\mathrm{4}}{\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}{b}+\mathrm{1}}\geqslant\frac{\mathrm{3}}{\mathrm{3}{a}+{b}+\mathrm{1}}+\frac{\mathrm{3}}{{a}+\mathrm{3}{b}+\mathrm{1}} \\ $$$$ \\ $$
Commented by HongKing last updated on 28/Nov/21
$$\mathrm{Cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$