Question Number 4809 by 314159 last updated on 15/Mar/16
$${Show}\:{that}\:\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:>\:\frac{{x}+{a}}{{x}−{a}}. \\ $$
Answered by Yozzii last updated on 15/Mar/16
$${Let}\:\phi=\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\frac{{x}+{a}}{{x}−{a}}.\:{Since}\:{x}^{\mathrm{2}} −{a}^{\mathrm{2}} =\left({x}−{a}\right)\left({x}+{a}\right) \\ $$ $$\Rightarrow\phi=\frac{\mathrm{1}}{\left({x}−{a}\right)}\left(\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}+{a}}−{x}+{a}\right) \\ $$ $$\phi=\frac{\mathrm{1}}{{x}−{a}}\left(\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{2}{ax}−{a}^{\mathrm{2}} }{{x}+{a}}\right) \\ $$ $$\phi=\frac{\mathrm{2}{ax}}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$ $${Let}\:{x}=\mathrm{1}\:{and}\:{a}=−\mathrm{2}.\Rightarrow\mathrm{2}{ax}=\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)<\mathrm{0} \\ $$ $${a}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{4}−\mathrm{1}=\mathrm{3}>\mathrm{0}.\:\therefore\:\phi<\mathrm{0}\Rightarrow\exists\left({x},{a}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$ $${such}\:{that}\:\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\ngtr\frac{{x}+{a}}{{x}−{a}}.\: \\ $$ $$\frac{\mathrm{1}+\mathrm{4}}{\mathrm{1}−\mathrm{4}}=\frac{\mathrm{5}}{−\mathrm{3}}\ngtr\frac{\mathrm{1}−\mathrm{2}}{\mathrm{1}−\left(−\mathrm{2}\right)}=\frac{−\mathrm{1}}{\mathrm{3}}\: \\ $$ $$\frac{−\mathrm{5}}{\mathrm{3}}<\frac{−\mathrm{1}}{\mathrm{3}} \\ $$ $$ \\ $$