Question Number 129235 by enter last updated on 14/Jan/21
$$\:\:\boldsymbol{{a}}=\:\sqrt[{\mathrm{3}}]{\mathrm{7}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\: \\ $$$$\:\:\:\boldsymbol{{b}}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Taqqoslang}} \\ $$
Commented by enter last updated on 14/Jan/21
$${compare} \\ $$
Commented by enter last updated on 14/Jan/21
$$??????? \\ $$
Answered by bemath last updated on 14/Jan/21
$$\:\sqrt[{\mathrm{3}}]{\mathrm{7}}\:+\sqrt[{\mathrm{3}}]{\mathrm{9}}\:=\:\frac{\mathrm{16}}{\:\sqrt[{\mathrm{3}}]{\mathrm{49}}−\sqrt[{\mathrm{3}}]{\mathrm{63}}+\sqrt[{\mathrm{3}}]{\mathrm{81}}}\:<\:\mathrm{4} \\ $$
Answered by mr W last updated on 13/Jun/21
![f(x)=(x)^(1/3) f′(x)=(x^(−(2/3)) /3)>0 for x>0 f′′(x)=−((2x^(−(5/3)) )/9)<0 for x>0 that means for x>0, f(x) is increasing, f′(x) is decreasing. f(7)=(7)^(1/3) =((8−1))^(1/3) =(8)^(1/3) −((f(8)−f(7))/1) f(9)=(9)^(1/3) =((8+1))^(1/3) =(8)^(1/3) +((f(9)−f(8))/1) ((f(9)−f(8))/1)<((f(8)−f(7))/1) f(7)+f(9)=2(8)^(1/3) +[((f(9)−f(8))/1)−((f(8)−f(7))/1)] <2(8)^(1/3) ⇒(7)^(1/3) +(9)^(1/3) <2(8)^(1/3) =2×2=4](https://www.tinkutara.com/question/Q143362.png)
$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{{x}} \\ $$$${f}'\left({x}\right)=\frac{{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}}>\mathrm{0}\:{for}\:{x}>\mathrm{0} \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{2}{x}^{−\frac{\mathrm{5}}{\mathrm{3}}} }{\mathrm{9}}<\mathrm{0}\:{for}\:{x}>\mathrm{0} \\ $$$${that}\:{means}\:{for}\:{x}>\mathrm{0},\:{f}\left({x}\right)\:{is}\:{increasing}, \\ $$$${f}'\left({x}\right)\:{is}\:{decreasing}. \\ $$$${f}\left(\mathrm{7}\right)=\sqrt[{\mathrm{3}}]{\mathrm{7}}=\sqrt[{\mathrm{3}}]{\mathrm{8}−\mathrm{1}}=\sqrt[{\mathrm{3}}]{\mathrm{8}}−\frac{{f}\left(\mathrm{8}\right)−{f}\left(\mathrm{7}\right)}{\mathrm{1}} \\ $$$${f}\left(\mathrm{9}\right)=\sqrt[{\mathrm{3}}]{\mathrm{9}}=\sqrt[{\mathrm{3}}]{\mathrm{8}+\mathrm{1}}=\sqrt[{\mathrm{3}}]{\mathrm{8}}+\frac{{f}\left(\mathrm{9}\right)−{f}\left(\mathrm{8}\right)}{\mathrm{1}} \\ $$$$\frac{{f}\left(\mathrm{9}\right)−{f}\left(\mathrm{8}\right)}{\mathrm{1}}<\frac{{f}\left(\mathrm{8}\right)−{f}\left(\mathrm{7}\right)}{\mathrm{1}} \\ $$$${f}\left(\mathrm{7}\right)+{f}\left(\mathrm{9}\right)=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{8}}+\left[\frac{{f}\left(\mathrm{9}\right)−{f}\left(\mathrm{8}\right)}{\mathrm{1}}−\frac{{f}\left(\mathrm{8}\right)−{f}\left(\mathrm{7}\right)}{\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{8}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{7}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}<\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{8}}=\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$