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Question Number 206355 by mr W last updated on 12/Apr/24
if the sum of three positive real   numbers is equal to their product,  prove that at least one of the   numbers is larger than 1.7.
$${if}\:{the}\:{sum}\:{of}\:{three}\:{positive}\:{real}\: \\ $$$${numbers}\:{is}\:{equal}\:{to}\:{their}\:{product}, \\ $$$${prove}\:{that}\:{at}\:{least}\:{one}\:{of}\:{the}\: \\ $$$${numbers}\:{is}\:{larger}\:{than}\:\mathrm{1}.\mathrm{7}. \\ $$
Answered by A5T last updated on 12/Apr/24
a+b+c=abc≥3((abc))^(1/3) ⇒(abc)≥3^(3/2)   One must be at least (3^(3/2) )^(1/3) =(√3)≈1.732
$${a}+{b}+{c}={abc}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\Rightarrow\left({abc}\right)\geqslant\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${One}\:{must}\:{be}\:{at}\:{least}\:\sqrt[{\mathrm{3}}]{\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} }=\sqrt{\mathrm{3}}\approx\mathrm{1}.\mathrm{732} \\ $$
Answered by mr W last updated on 12/Apr/24
s=a+b+c≥3((abc))^(1/3) =3((a+b+c))^(1/3) =3(s)^(1/3)   s^(2/3) ≥3 ⇒s≥(√3^3 )=(√(27))  at least one number must be   ≥(s/3)≥(√3)>1.7
$${s}={a}+{b}+{c}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}=\mathrm{3}\sqrt[{\mathrm{3}}]{{a}+{b}+{c}}=\mathrm{3}\sqrt[{\mathrm{3}}]{{s}} \\ $$$${s}^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{3}\:\Rightarrow{s}\geqslant\sqrt{\mathrm{3}^{\mathrm{3}} }=\sqrt{\mathrm{27}} \\ $$$${at}\:{least}\:{one}\:{number}\:{must}\:{be}\: \\ $$$$\geqslant\frac{{s}}{\mathrm{3}}\geqslant\sqrt{\mathrm{3}}>\mathrm{1}.\mathrm{7} \\ $$

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