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Question Number 12040 by Mr Chheang Chantria last updated on 10/Apr/17

Prove that ∀x,y∈R ⇒ 1+ ((x^2 +y^2 +xy)/3) ≥ x+y

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\forall\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{R}} \\ $$$$\Rightarrow\:\mathrm{1}+\:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}\:\geqslant\:\boldsymbol{{x}}+\boldsymbol{{y}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17

x^2 +y^2 +xy≥3x+3y−3⇒ x^2 −2x+1+y^2 −2y+1+xy−x−y+1≥0 (x−1)^2 +(y−1)^2 +x(y−1)−(y−1)≥0 (x−1)^2 +(y−1)^2 +(x−1)(y−1)≥0 this iniquality,is valid for evrey x&y∈R.■ (A^2 +B^2 +AB≥0 ∀A,B∈R.)

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}\geqslant\mathrm{3}{x}+\mathrm{3}{y}−\mathrm{3}\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}+{xy}−{x}−{y}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +{x}\left({y}−\mathrm{1}\right)−\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$${this}\:{iniquality},{is}\:{valid}\:{for}\:{evrey}\:{x\&y}\in{R}.\blacksquare \\ $$$$\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{AB}\geqslant\mathrm{0}\:\:\:\forall{A},{B}\in\boldsymbol{{R}}.\right) \\ $$

Commented by Mr Chheang Chantria last updated on 11/Apr/17

Nice solution. Thanks you

$$\boldsymbol{{Nice}}\:\boldsymbol{{solution}}.\:\boldsymbol{{Thanks}}\:\boldsymbol{{you}} \\ $$

Answered by ajfour last updated on 10/Apr/17

let x+y=u and x−y=v ⇒ xy=(1/4)(u^2 −v^2 ) consider : f(x,y)=1+((x^2 +y^2 +xy)/3)−(x+y) = 1+(((x+y)^2 −xy)/3) − (x+y) = 1+(1/3){ u^2 −(1/4)(u^2 −v^2 )}−u =1+ (1/3){(3/4)u^2 +(v^2 /4) }−u = (u^2 /4)−u+1+(v^2 /(12)) = ((u/2)−1)^2 +(v^2 /(12)) ≥ 0 or 1+((x^2 +y^2 +xy)/3)−(x+y) = (((x+y)/2)−1)^2 +(((x−y)^2 )/(12)) ≥ 0 .

$${let}\:\:\:{x}+{y}={u}\:\:{and}\:\:{x}−{y}={v} \\ $$$$\Rightarrow\:{xy}=\frac{\mathrm{1}}{\mathrm{4}}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right) \\ $$$${consider}\:: \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}−\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:=\:\mathrm{1}+\frac{\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)^{\mathrm{2}} −\boldsymbol{{xy}}}{\mathrm{3}}\:−\:\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left\{\:\boldsymbol{{u}}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left(\boldsymbol{{u}}^{\mathrm{2}} −\boldsymbol{{v}}^{\mathrm{2}} \right)\right\}−\boldsymbol{{u}} \\ $$$$\:\:\:\:=\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{{u}}^{\mathrm{2}} +\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{4}}\:\right\}−\boldsymbol{{u}} \\ $$$$\:\:\:\:=\:\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{u}}+\mathrm{1}+\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\:\:\:\:=\:\left(\frac{\boldsymbol{{u}}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\boldsymbol{{v}}^{\mathrm{2}} }{\mathrm{12}}\:\:\geqslant\:\mathrm{0} \\ $$$${or} \\ $$$$\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}−\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:\:\:\:=\:\left(\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)^{\mathrm{2}} }{\mathrm{12}}\:\geqslant\:\mathrm{0}\:. \\ $$

Commented by Mr Chheang Chantria last updated on 11/Apr/17

Good Solution. Thanks you

$$\boldsymbol{{Good}}\:\boldsymbol{{Solution}}.\:\boldsymbol{{Thanks}}\:\boldsymbol{{you}} \\ $$

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