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Question Number 118810 by greg_ed last updated on 19/Oct/20

(√(2003)) + (√(2005)) < 2(√(2004))  ???

$$\sqrt{\mathrm{2003}}\:+\:\sqrt{\mathrm{2005}}\:<\:\mathrm{2}\sqrt{\mathrm{2004}}\:\:??? \\ $$

Answered by Olaf last updated on 19/Oct/20

  x = (√(2003))  x = (√(2004−1))  x = (√(2004))(√(1−(1/(2004))))  x = (√(2004))(1−(1/2).(1/(2004))−(1/8).(1/(2004^2 ))...)    y = (√(2005))  y = (√(2004+1))  y = (√(2004))(√(1+(1/(2004))))  y = (√(2004))(1+(1/2).(1/(2004))−(1/8).(1/(2004^2 ))...)    x+y = 2(√(2004))−(1/(4.2004^(3/2) ))... ≤ 2(√(2004))

$$ \\ $$ $${x}\:=\:\sqrt{\mathrm{2003}} \\ $$ $${x}\:=\:\sqrt{\mathrm{2004}−\mathrm{1}} \\ $$ $${x}\:=\:\sqrt{\mathrm{2004}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2004}}} \\ $$ $${x}\:=\:\sqrt{\mathrm{2004}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2004}}−\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{1}}{\mathrm{2004}^{\mathrm{2}} }...\right) \\ $$ $$ \\ $$ $${y}\:=\:\sqrt{\mathrm{2005}} \\ $$ $${y}\:=\:\sqrt{\mathrm{2004}+\mathrm{1}} \\ $$ $${y}\:=\:\sqrt{\mathrm{2004}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2004}}} \\ $$ $${y}\:=\:\sqrt{\mathrm{2004}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2004}}−\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{1}}{\mathrm{2004}^{\mathrm{2}} }...\right) \\ $$ $$ \\ $$ $${x}+{y}\:=\:\mathrm{2}\sqrt{\mathrm{2004}}−\frac{\mathrm{1}}{\mathrm{4}.\mathrm{2004}^{\mathrm{3}/\mathrm{2}} }...\:\leqslant\:\mathrm{2}\sqrt{\mathrm{2004}} \\ $$

Commented bygreg_ed last updated on 20/Oct/20

thanks for your help !

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}\:! \\ $$

Answered by MJS_new last updated on 19/Oct/20

(√(a−1))+(√(a+1))<2(√a)  defined for a≥1  both sides >0 ⇒ we are allowed to square  a−1+2(√(a−1))(√(a+1))+a+1<4a  2a+2(√(a−1))(√(a+1))<4a  (√(a−1))(√(a+1))<a  squaring again  a^2 −1<a^2   true ∀a≥1

$$\sqrt{{a}−\mathrm{1}}+\sqrt{{a}+\mathrm{1}}<\mathrm{2}\sqrt{{a}} \\ $$ $$\mathrm{defined}\:\mathrm{for}\:{a}\geqslant\mathrm{1} \\ $$ $$\mathrm{both}\:\mathrm{sides}\:>\mathrm{0}\:\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$ $${a}−\mathrm{1}+\mathrm{2}\sqrt{{a}−\mathrm{1}}\sqrt{{a}+\mathrm{1}}+{a}+\mathrm{1}<\mathrm{4}{a} \\ $$ $$\mathrm{2}{a}+\mathrm{2}\sqrt{{a}−\mathrm{1}}\sqrt{{a}+\mathrm{1}}<\mathrm{4}{a} \\ $$ $$\sqrt{{a}−\mathrm{1}}\sqrt{{a}+\mathrm{1}}<{a} \\ $$ $$\mathrm{squaring}\:\mathrm{again} \\ $$ $${a}^{\mathrm{2}} −\mathrm{1}<{a}^{\mathrm{2}} \\ $$ $$\mathrm{true}\:\forall{a}\geqslant\mathrm{1} \\ $$

Commented bygreg_ed last updated on 20/Oct/20

thanks for your help !

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}\:! \\ $$

Answered by 1549442205PVT last updated on 20/Oct/20

Apply the inequality(x+y)<(√(2(x^2 +y^2 )))   for ∀ x,y>0 x≠y we have  (√(2003))+(√(2005))<(√(2(2003+2005)))  =(√(4.2004))=2(√(2004))

$$\mathrm{Apply}\:\mathrm{the}\:\mathrm{inequality}\left(\mathrm{x}+\mathrm{y}\right)<\sqrt{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}\: \\ $$ $$\mathrm{for}\:\forall\:\mathrm{x},\mathrm{y}>\mathrm{0}\:\mathrm{x}\neq\mathrm{y}\:\mathrm{we}\:\mathrm{have} \\ $$ $$\sqrt{\mathrm{2003}}+\sqrt{\mathrm{2005}}<\sqrt{\mathrm{2}\left(\mathrm{2003}+\mathrm{2005}\right)} \\ $$ $$=\sqrt{\mathrm{4}.\mathrm{2004}}=\mathrm{2}\sqrt{\mathrm{2004}} \\ $$

Commented bygreg_ed last updated on 20/Oct/20

thanks for your help !

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}\:! \\ $$

Commented by1549442205PVT last updated on 25/Oct/20

You are welcome

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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