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Question Number 149585 by mathdanisur last updated on 06/Aug/21
if  x;y;z>0  and  x^2 +y^2 +z^2 =3  then:  Σ ((x^2 +y^2 )/((2x^2 +y^2 )(y^2 +2x^2 ))) ≥ (2/3)
$${if}\:\:{x};{y};{z}>\mathrm{0}\:\:{and}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{3}\:\:{then}: \\ $$$$\Sigma\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\left(\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \right)}\:\geqslant\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by dumitrel last updated on 06/Aug/21
(√((2x^2 +y^2 )(2y^2 +x^2 )))≤^(am−gm) ((3(x^2 +y^2 ))/2)⇒  (2x^2 +y^2 )(2y^2 +x^2 )≤((9(x^2 +y^2 )^2 )/4)  Σ((x^2 +y^2 )/((2x^2 +y^2 )(2y^2 +x^2 )))≥Σ((x^2 +y^2 )/((9(x^2 +y^2 )^2 )/4))=  (4/9)Σ(1/(x^2 +y^2 ))≥(4/9)∙(9/(2(x^2 +y^2 +z^2 )))=(2/3)
$$\sqrt{\left(\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left(\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\overset{{am}−{gm}} {\leqslant}\frac{\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\mathrm{2}}\Rightarrow \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left(\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\leqslant\frac{\mathrm{9}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Sigma\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\left(\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left(\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\geqslant\Sigma\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\frac{\mathrm{9}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}}= \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\Sigma\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\geqslant\frac{\mathrm{4}}{\mathrm{9}}\centerdot\frac{\mathrm{9}}{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 06/Aug/21
Thank you Ser, cool
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{cool} \\ $$

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