x;y;z>0, γ≥0, x^3 +y^3 +z^3 +xyz=4 proof: (x+y+z)^3 +γ(x^3 +y^3 +z^3 )≥27+3γ


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Question Number 139762 by mathdanisur last updated on 01/May/21

$x; y; z > 0, γ ⩾ 0, x^{3} + y^{3} + z^{3} + x y z = 4$ $p r o o f : {(x + y + z)}^{3} + γ (x^{3} + y^{3} + z^{3}) ⩾ 27 + 3 γ$
	Answered by mindispower last updated on 02/May/21

	$A M - G M x^{3} + y^{3} + z^{3} ⩾ 3 \sqrt[3]{x^{3} y^{3} z^{3}} = 3 x y z$ $\Rightarrow 4 ⩾ 4 x y z \Rightarrow x^{3} + y^{3} + z^{3} = 4 - x y z ⩾ 3$ $(x + y + z) ⩾ 3 \sqrt[3]{x y z} \Rightarrow {(x + y + z)}^{3} ⩾ 27 x y z ⩾ 27$ ${(x + y + z)}^{3} + γ (x^{3} + y^{3} + z^{3}) ⩾ 27 + 3 γ$
	Commented bymathdanisur last updated on 02/May/21

	$27 x y z ⩽ 27, s i n c e x y z ⩽ 1,$ $p l e a s e c h e c k s o l u t i o n s i r$
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