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Question Number 4809 by 314159 last updated on 15/Mar/16
Show that ((x^2 +a^2 )/(x^2 −a^2 )) > ((x+a)/(x−a)).
Showthatx2+a2x2−a2>x+ax−a.
Answered by Yozzii last updated on 15/Mar/16
Let φ=((x^2 +a^2 )/(x^2 −a^2 ))−((x+a)/(x−a)). Since x^2 −a^2 =(x−a)(x+a) ⇒φ=(1/((x−a)))(((x^2 +a^2 )/(x+a))−x+a) φ=(1/(x−a))(((x^2 +a^2 −x^2 −2ax−a^2 )/(x+a))) φ=((2ax)/(a^2 −x^2 )) Let x=1 and a=−2.⇒2ax=2×1×(−1)<0 a^2 −x^2 =4−1=3>0. ∴ φ<0⇒∃(x,a)∈R^2 such that ((x^2 +a^2 )/(x^2 −a^2 ))≯((x+a)/(x−a)). ((1+4)/(1−4))=(5/(−3))≯((1−2)/(1−(−2)))=((−1)/3) ((−5)/3)<((−1)/3)
Letϕ=x2+a2x2−a2−x+ax−a.Sincex2−a2=(x−a)(x+a)⇒ϕ=1(x−a)(x2+a2x+a−x+a)ϕ=1x−a(x2+a2−x2−2ax−a2x+a)ϕ=2axa2−x2Letx=1anda=−2.⇒2ax=2×1×(−1)<0a2−x2=4−1=3>0.∴ϕ<0⇒∃(x,a)∈R2suchthatx2+a2x2−a2≯x+ax−a.1+41−4=5−3≯1−21−(−2)=−13−53<−13

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