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Question Number 79147 by jagoll last updated on 23/Jan/20

(√(x+(1/x^2 )))+(√(x−(1/x^2 ) ))≤(2/x)

x+1x2+x−1x2⩽2x

Answered by john santu last updated on 23/Jan/20

(√((x^3 +1)/x^2 ))+(√((x^3 −1)/x^2 ))≤(2/x) (i)x≥−1 ∧x≥1∧x≠0 (ii) (((√(x^3 +1))+(√(x^3 −1)))/x)≤(2/x) ⇒(√(x^3 +1))+(√(x^3 −1))≤2 let x^3 =t⇒(√(t+1))+(√(t−1)) ≤2 2t+2(√(t^2 −1))≤4 (√(t^2 −1)) ≤2−t⇒t^2 −1≤4−4t+t^2 4t≤5 ⇒t≤(5/4) ⇒x^3 ≤(5/4) x≤((5/4))^(1/(3 )) . now we get solution x≥1 ∧x≤((5/4))^(1/(3 )) ⇒ x∈[1,((5/4))^(1/(3 )) ]

x3+1x2+x3−1x2⩽2x(i)x⩾−1∧x⩾1∧x≠0(ii)x3+1+x3−1x⩽2x⇒x3+1+x3−1⩽2letx3=t⇒t+1+t−1⩽22t+2t2−1⩽4t2−1⩽2−t⇒t2−1⩽4−4t+t24t⩽5⇒t⩽54⇒x3⩽54x⩽543.nowwegetsolutionx⩾1∧x⩽543⇒x∈[1,543]

Commented by jagoll last updated on 23/Jan/20

thank you sir. your solution is fantastic

thankyousir.yoursolutionisfantastic

Commented by jagoll last updated on 23/Jan/20

good sir

goodsir

Commented by MJS last updated on 23/Jan/20

nice and right, but (just saying): (i) x≥−1∧x≥1∧x≠0 ⇒ x≥1 squaring without substituting t isn′t any harder: 0≤(√(x^3 +1))+(√(x^3 −1))≤2 ∣^2 0≤2x^3 +2(√(x^3 +1))(√(x^3 −1))≤4 −x^3 ≤(√(x^3 +1))(√(x^3 −1))≤2−x^3 but we know that 0≤(√(x^3 +1))(√(x^3 −1)) ⇒ 0≤4−x^3 ⇒ x≤(2)^(1/3) 0≤(√(x^3 +1))(√(x^3 −1))≤2−x^3 ∣^2 0≤x^6 −1≤x^6 −4x^3 +4 0≤x^3 ≤(5/4) 0≤x≤((5/4))^(1/3) but x≥1 ⇒ 1≤x≤((5/4))^(1/3)

niceandright,but(justsaying):(i)x⩾−1∧x⩾1∧x≠0⇒x⩾1squaringwithoutsubstitutingtisn′tanyharder:0⩽x3+1+x3−1⩽2∣20⩽2x3+2x3+1x3−1⩽4−x3⩽x3+1x3−1⩽2−x3butweknowthat0⩽x3+1x3−1⇒0⩽4−x3⇒x⩽230⩽x3+1x3−1⩽2−x3∣20⩽x6−1⩽x6−4x3+40⩽x3⩽540⩽x⩽543butx⩾1⇒1⩽x⩽543

Commented by john santu last updated on 23/Jan/20

yes depending on the direction of our view sir

yesdependingonthedirectionofourviewsir

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