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

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