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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)).
$${Show}\:{that}\:\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:>\:\frac{{x}+{a}}{{x}−{a}}. \\ $$
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)
$${Let}\:\phi=\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\frac{{x}+{a}}{{x}−{a}}.\:{Since}\:{x}^{\mathrm{2}} −{a}^{\mathrm{2}} =\left({x}−{a}\right)\left({x}+{a}\right) \\ $$$$\Rightarrow\phi=\frac{\mathrm{1}}{\left({x}−{a}\right)}\left(\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}+{a}}−{x}+{a}\right) \\ $$$$\phi=\frac{\mathrm{1}}{{x}−{a}}\left(\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{2}{ax}−{a}^{\mathrm{2}} }{{x}+{a}}\right) \\ $$$$\phi=\frac{\mathrm{2}{ax}}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${Let}\:{x}=\mathrm{1}\:{and}\:{a}=−\mathrm{2}.\Rightarrow\mathrm{2}{ax}=\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)<\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{4}−\mathrm{1}=\mathrm{3}>\mathrm{0}.\:\therefore\:\phi<\mathrm{0}\Rightarrow\exists\left({x},{a}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$${such}\:{that}\:\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\ngtr\frac{{x}+{a}}{{x}−{a}}.\: \\ $$$$\frac{\mathrm{1}+\mathrm{4}}{\mathrm{1}−\mathrm{4}}=\frac{\mathrm{5}}{−\mathrm{3}}\ngtr\frac{\mathrm{1}−\mathrm{2}}{\mathrm{1}−\left(−\mathrm{2}\right)}=\frac{−\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{−\mathrm{5}}{\mathrm{3}}<\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$

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