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

$${x};{y};{z}>\mathrm{0},\:\gamma\geqslant\mathrm{0},\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +{xyz}=\mathrm{4} \\ $$$${proof}:\:\left({x}+{y}+{z}\right)^{\mathrm{3}} +\gamma\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\geqslant\mathrm{27}+\mathrm{3}\gamma \\ $$

Answered by mindispower last updated on 02/May/21

AM−GM   x^3 +y^3 +z^3 ≥3((x^3 y^3 z^3 ))^(1/3) =3xyz  ⇒4≥4xyz⇒x^3 +y^3 +z^3 =4−xyz≥3  (x+y+z)≥3((xyz))^(1/3) ⇒(x+y+z)^3 ≥27xyz≥27  (x+y+z)^3 +γ(x^3 +y^3 +z^3 )≥27+3γ

$${AM}−{GM}\:\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} {y}^{\mathrm{3}} {z}^{\mathrm{3}} }=\mathrm{3}{xyz} \\ $$$$\Rightarrow\mathrm{4}\geqslant\mathrm{4}{xyz}\Rightarrow{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{4}−{xyz}\geqslant\mathrm{3} \\ $$$$\left({x}+{y}+{z}\right)\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{xyz}}\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{3}} \geqslant\mathrm{27}{xyz}\geqslant\mathrm{27} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} +\gamma\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\geqslant\mathrm{27}+\mathrm{3}\gamma \\ $$

Commented by mathdanisur last updated on 02/May/21

27xyz≤27, since xyz≤1,  please check solution sir

$$\mathrm{27}{xyz}\leqslant\mathrm{27},\:{since}\:{xyz}\leqslant\mathrm{1}, \\ $$$${please}\:{check}\:{solution}\:{sir} \\ $$

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