Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 175396 by Shrinava last updated on 29/Aug/22

Answered by mahdipoor last updated on 29/Aug/22

x+y^2 (√(2/(x^2 +y^2 )))≥y+(√((x^2 +y^2 )/2)) ⇔ (x+y^2 (√(2/(x^2 +y^2 ))))((√((x^2 +y^2 )/2)))≥(y+(√((x^2 +y^2 )/2)))((√((x^2 +y^2 )/2))) ⇔ x(√((x^2 +y^2 )/2))+y^2 ≥y(√((x^2 +y^2 )/2))+((x^2 +y^2 )/2) ⇔ (x−y)(√((x^2 +y^2 )/2))≥((x^2 −y^2 )/2)=(((x−y)(x+y))/2) ⇔^I (√(((x^2 +y^2 )/2)≥))((x+y)/2) ⇔^(II) ((x^2 +y^2 )/2)≥((x^2 +y^2 +2xy)/4) ⇔ ((x^2 +y^2 −2xy)/4)=(((x−y)^2 )/4)≥0 I : x−y≥0 or x≥y II : x + y≥0

$${x}+{y}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\geqslant{y}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\:\:\:\Leftrightarrow \\ $$$$\left({x}+{y}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\right)\left(\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\right)\geqslant\left({y}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\right)\left(\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\right)\:\:\:\Leftrightarrow \\ $$$${x}\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}+{y}^{\mathrm{2}} \geqslant{y}\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}+\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\Leftrightarrow \\ $$$$\left({x}−{y}\right)\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\geqslant\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left({x}−{y}\right)\left({x}+{y}\right)}{\mathrm{2}}\:\:\:\overset{\mathrm{I}} {\Leftrightarrow} \\ $$$$\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\geqslant}\frac{{x}+{y}}{\mathrm{2}}\:\overset{\mathrm{II}} {\Leftrightarrow}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\geqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}}{\mathrm{4}}\:\Leftrightarrow \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}{\mathrm{4}}=\frac{\left({x}−{y}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant\mathrm{0}\:\: \\ $$$$\mathrm{I}\::\:{x}−{y}\geqslant\mathrm{0}\:\:\:{or}\:\:{x}\geqslant{y} \\ $$$$\mathrm{II}\::\:\:{x}\:+\:{y}\geqslant\mathrm{0}\:\: \\ $$

Answered by MJS_new last updated on 29/Aug/22

let y=px with p>0 x(((√2)p^2 +(√(p^2 +1)))/( (√(p^2 +1))))≥x(((√2)p+(√(p^2 +1)))/( (√2))) (((√2)p^2 +(√(p^2 +1)))/( (√(p^2 +1))))≥(((√2)p+(√(p^2 +1)))/( (√2))) 2p^2 +(√(2(p^2 +1)))≥p^2 +1+p(√(2(p^2 +1))) (p−1)(p+1−(√(2(p^2 +1))))≥0 only true for 0<p≤1 ⇒ only true for 0<y≤x

$$\mathrm{let}\:{y}={px}\:\mathrm{with}\:{p}>\mathrm{0} \\ $$$${x}\frac{\sqrt{\mathrm{2}}{p}^{\mathrm{2}} +\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}\geqslant{x}\frac{\sqrt{\mathrm{2}}{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\sqrt{\mathrm{2}}{p}^{\mathrm{2}} +\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}\geqslant\frac{\sqrt{\mathrm{2}}{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{1}\right)}\geqslant{p}^{\mathrm{2}} +\mathrm{1}+{p}\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\left({p}−\mathrm{1}\right)\left({p}+\mathrm{1}−\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{1}\right)}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{0}<{p}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{0}<{y}\leqslant{x} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com