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Question Number 147582 by mathdanisur last updated on 22/Jul/21

if  x;y;z>0  prove that  ((x^3 +y^3 +z^3 )/(xyz)) ≥ 2((x/(y+z)) + (y/(z+x)) + (z/(x+y)))

$${if}\:\:{x};{y};{z}>\mathrm{0}\:\:{prove}\:{that} \\ $$ $$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} }{{xyz}}\:\geqslant\:\mathrm{2}\left(\frac{{x}}{{y}+{z}}\:+\:\frac{{y}}{{z}+{x}}\:+\:\frac{{z}}{{x}+{y}}\right) \\ $$

Answered by mitica last updated on 22/Jul/21

m_h ≤m_a ⇒(1/(a+b))≤((a+b)/(4ab))  2Σ(x/(y+z))≤2Σ((x(y+z))/(4yz))=Σ((x^2 (y+z))/(2xyz))=  (1/(2xyz))Σxy(x+y)≤(1/(2xyz))Σ(x^3 +y^3 )=((2(x^3 +y^3 +z^3 ))/(2xyz))=((x^3 +y^3 +z^3 )/(xyz))

$${m}_{{h}} \leqslant{m}_{{a}} \Rightarrow\frac{\mathrm{1}}{{a}+{b}}\leqslant\frac{{a}+{b}}{\mathrm{4}{ab}} \\ $$ $$\mathrm{2}\Sigma\frac{{x}}{{y}+{z}}\leqslant\mathrm{2}\Sigma\frac{{x}\left({y}+{z}\right)}{\mathrm{4}{yz}}=\Sigma\frac{{x}^{\mathrm{2}} \left({y}+{z}\right)}{\mathrm{2}{xyz}}= \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}{xyz}}\Sigma{xy}\left({x}+{y}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2}{xyz}}\Sigma\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)=\frac{\mathrm{2}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)}{\mathrm{2}{xyz}}=\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} }{{xyz}} \\ $$

Commented bymathdanisur last updated on 22/Jul/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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