Question Number 155481 by mathdanisur last updated on 01/Oct/21
$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\in\left[\mathrm{1};\infty\right) \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{a}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}} \:;\:\mathrm{b}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}}} \:;\:\mathrm{c}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{c}}}} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}. \\ $$
Answered by mr W last updated on 01/Oct/21
$${for}\:{x}\in\left[\mathrm{1},\infty\right): \\ $$$$\mathrm{1}\leqslant{x}^{\frac{\mathrm{1}}{{x}}} \leqslant{e}^{\frac{\mathrm{1}}{{e}}} \approx\mathrm{1}.\mathrm{4447} \\ $$$${that}\:{means}\:\mathrm{1}\leqslant{a}^{\frac{\mathrm{1}}{{a}}} ,{b}^{\frac{\mathrm{1}}{{b}}} ,{c}^{\frac{\mathrm{1}}{{c}}} \leqslant{e}^{\frac{\mathrm{1}}{{e}}} . \\ $$$${say}\:{a}^{\frac{\mathrm{1}}{{a}}} \leqslant{b}^{\frac{\mathrm{1}}{{b}}} \leqslant{c}^{\frac{\mathrm{1}}{{c}}} , \\ $$$${c}^{\frac{\mathrm{1}}{{c}}} \leqslant{e}^{\frac{\mathrm{1}}{{e}}} \approx\mathrm{1}.\mathrm{4447} \\ $$$${a}^{\frac{\mathrm{1}}{{a}}} +{b}^{\frac{\mathrm{1}}{{b}}} \geqslant\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\Rightarrow{a}^{\frac{\mathrm{1}}{{a}}} +{b}^{\frac{\mathrm{1}}{{b}}} >{c}^{\frac{\mathrm{1}}{{c}}} \\ $$$${i}.{e}.\:{a}^{\frac{\mathrm{1}}{{a}}} ,{b}^{\frac{\mathrm{1}}{{b}}} ,{c}^{\frac{\mathrm{1}}{{c}}} \:{can}\:{form}\:{a}\:{triangle}. \\ $$
Commented by mathdanisur last updated on 01/Oct/21
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$