Question Number 12040 by Mr Chheang Chantria last updated on 10/Apr/17
$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\forall\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{R}} \\ $$$$\Rightarrow\:\mathrm{1}+\:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}\:\geqslant\:\boldsymbol{{x}}+\boldsymbol{{y}} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}\geqslant\mathrm{3}{x}+\mathrm{3}{y}−\mathrm{3}\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}+{xy}−{x}−{y}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +{x}\left({y}−\mathrm{1}\right)−\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$${this}\:{iniquality},{is}\:{valid}\:{for}\:{evrey}\:{x\&y}\in{R}.\blacksquare \\ $$$$\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{AB}\geqslant\mathrm{0}\:\:\:\forall{A},{B}\in\boldsymbol{{R}}.\right) \\ $$
Commented by Mr Chheang Chantria last updated on 11/Apr/17
$$\boldsymbol{{Nice}}\:\boldsymbol{{solution}}.\:\boldsymbol{{Thanks}}\:\boldsymbol{{you}} \\ $$
Answered by ajfour last updated on 10/Apr/17
$${let}\:\:\:{x}+{y}={u}\:\:{and}\:\:{x}−{y}={v} \\ $$$$\Rightarrow\:{xy}=\frac{\mathrm{1}}{\mathrm{4}}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right) \\ $$$${consider}\:: \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}−\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:=\:\mathrm{1}+\frac{\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)^{\mathrm{2}} −\boldsymbol{{xy}}}{\mathrm{3}}\:−\:\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left\{\:\boldsymbol{{u}}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left(\boldsymbol{{u}}^{\mathrm{2}} −\boldsymbol{{v}}^{\mathrm{2}} \right)\right\}−\boldsymbol{{u}} \\ $$$$\:\:\:\:=\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{{u}}^{\mathrm{2}} +\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{4}}\:\right\}−\boldsymbol{{u}} \\ $$$$\:\:\:\:=\:\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{u}}+\mathrm{1}+\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\:\:\:\:=\:\left(\frac{\boldsymbol{{u}}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{12}}\:\:\geqslant\:\mathrm{0} \\ $$$${or} \\ $$$$\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}−\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:\:\:\:=\:\left(\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)^{\mathrm{2}} }{\mathrm{12}}\:\geqslant\:\mathrm{0}\:. \\ $$
Commented by Mr Chheang Chantria last updated on 11/Apr/17
$$\boldsymbol{{Good}}\:\boldsymbol{{Solution}}.\:\boldsymbol{{Thanks}}\:\boldsymbol{{you}} \\ $$