Question-175396

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

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

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