prove a^2 +b^2 +c^2 +((abc))^(1/3) ≥4 if ab+bc+ac=3


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Question Number 204948 by TonyCWX08 last updated on 03/Mar/24

$p r o v e$ $a^{2} + b^{2} + c^{2} + \sqrt[3]{a b c} ⩾ 4$ $i f$ $a b + b c + a c = 3$
Commented by TonyCWX08 last updated on 04/Mar/24

$C a n a n y o n e p l e a s e h e l p ? ? ?$
Commented by mr W last updated on 04/Mar/24

$q u e s t i o n i s w r o n g!$ $w i t h a = b = c = - 1 y o u h a v e$ $a b + b c + c a = 3$ $b u t a^{2} + b^{2} + c^{2} + \sqrt[3]{a b c} = 1 + 1 + 1 - 1 = 2 ⩾ 4$
Commented by mr W last updated on 04/Mar/24

$m a y b e y o u s h o u l d s t a t e a, b, c \in R^{+}$
Commented by TonyCWX08 last updated on 05/Mar/24

$T h e y A r e P o s i t i v e R e a l N u m b e r s . . . . .$
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