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Question Number 206862 by hardmath last updated on 28/Apr/24

Commented by SWPlaysMC last updated on 29/Apr/24

$$\mathrm{Easiest}\mathrm{way}\mathrm{I}\mathrm{thought}\mathrm{of}\mathrm{proving}\mathrm{this}\left(\mathrm{I}\mathrm{partially}\mathrm{solved}\mathrm{this}\right)$$$$\mathrm{Let}a=1,b=1,\mathrm{and}c=1$$$$(\mathrm{Hint}:\mathrm{if}\mathrm{any}\mathrm{one}\mathrm{of}\mathrm{these}\mathrm{is}0\mathrm{then}\mathrm{expression}\mathrm{is}\mathrm{theoretically}\mathrm{undefined}\mathrm{due}\mathrm{to}\mathbf{÷}\mathrm{by}0)$$$$\frac{\sqrt{\frac{1}{1}}+\frac{1}{1}}{\sqrt{\frac{1}{1}}+\sqrt{\frac{1}{1}}}+\frac{\sqrt{\frac{1}{1}}+\frac{1}{1}}{\sqrt{\frac{1}{1}}+\sqrt{\frac{1}{1}}}+\frac{\sqrt{\frac{1}{1}}+\frac{1}{1}}{\sqrt{\frac{1}{1}}+\sqrt{\frac{1}{1}}}⩾3$$$$\mathrm{so}\mathrm{all}\sqrt{\frac{1}{1}}=1\mathrm{and}\frac{1}{1}=1,\mathrm{thus}\mathrm{substitute}\mathrm{all}\mathrm{nums}\mathrm{and}\mathrm{denoms}\mathrm{with}1+1\mathrm{like}\mathrm{so}$$$$\frac{1+1}{1+1}+\frac{1+1}{1+1}+\frac{1+1}{1+1}⩾3$$$$1+1=2\mathrm{and}\frac{2}{2}=1\mathrm{so}$$$$1+1+1⩾3$$$$3⩾3\mathrm{or}3=3✓$$$$\mathrm{So}a,b,\mathrm{and}c\mathrm{all}\mathrm{have}\mathrm{to}\mathrm{be}\mathrm{at}\mathrm{least}⩾1\mathrm{or}\mathrm{the}\mathrm{expression}\mathrm{will}\mathrm{NOT}\mathrm{be}⩾3$$$$\mathrm{or}a⩾1,b⩾1,c⩾1,a+b+c\nless 3$$$$---------------$$$$\mathrm{Work}\mathrm{smarter},\mathrm{not}\mathrm{harder};\mathrm{p}$$

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