Question Number 204948 by TonyCWX08 last updated on 03/Mar/24
$${prove}\: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{{abc}}\geqslant\mathrm{4} \\ $$$${if} \\ $$$${ab}+{bc}+{ac}=\mathrm{3} \\ $$
Commented by TonyCWX08 last updated on 04/Mar/24
$$ \\ $$$${Can}\:{anyone}\:{please}\:{help}??? \\ $$
Commented by mr W last updated on 04/Mar/24
$${question}\:{is}\:{wrong}! \\ $$$${with}\:{a}={b}={c}=−\mathrm{1}\:{you}\:{have} \\ $$$${ab}+{bc}+{ca}=\mathrm{3} \\ $$$${but}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{{abc}}=\mathrm{1}+\mathrm{1}+\mathrm{1}−\mathrm{1}=\mathrm{2}\:\cancel{\geqslant}\:\mathrm{4} \\ $$
Commented by mr W last updated on 04/Mar/24
$${maybe}\:{you}\:{should}\:{state}\:{a},{b},{c}\:\in{R}^{+} \\ $$
Commented by TonyCWX08 last updated on 05/Mar/24
$${They}\:{Are}\:{Positive}\:{Real}\:{Numbers}….. \\ $$