Given a closed trajectory $x^4 - x^2 + y^4 - y^2= 0$ Prove that any distance between any point on the curve and the origin does not exceed $\sqrt2$ (ie, maximum distance from the origin to the curve is $\sqrt2$)
I proved it using polar transformations but i'd rather use proof by contradiction without using polar transformation, so i tried
$$\begin{array}{l} \left( {\exists \alpha ,\beta } \right){\rm{ }}{\alpha ^2} + {\beta ^2}>2\\ {\alpha ^4} - {\alpha ^2} + {\beta ^4} - {\beta ^2} = 0\\ \Rightarrow {\alpha ^4} + {\beta ^4} = {\alpha ^2} + {\beta ^2}\\ \Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2} = {\alpha ^2} + {\beta ^2}{\rm{ }}\left( {{\rm{Completing - Square}}} \right)\\ \Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} = {\alpha ^2} + {\beta ^2} + 2{\left( {\alpha \beta } \right)^2}\\ \Rightarrow {\left( {a + b} \right)^2} = a + b + 2ab \le 2\left( {{a^2} + {b^2}} \right){\rm{ }}\left( {{\rm{Cauchy - Schwarz - Inequality}}} \right)\\ \Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + {b^2}} \right) - 2ab\\ \Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + {\left( {a - b} \right)^2}\\ \Rightarrow {\left( {a - b} \right)^2} \ge \left( {a + b} \right) - \left( {{a^2} + {b^2}} \right) \ge 0\\ \Rightarrow \left( {a + b} \right) \ge \left( {{a^2} + {b^2}} \right)\\ \Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) \ge \left( {{\alpha ^4} + {\beta ^4}} \right)\\ \Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) - \left( {{\alpha ^4} + {\beta ^4}} \right) \ge 0 \end{array} % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGceaqabeaadaqadaqaaiabgoGiKiabeg7aHj % aacYcacqaHYoGyaiaawIcacaGLPaaacaqGGaGaeqySde2aaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaeqOSdi2aaWbaaSqabeaacaaIYaaaaO % GaaGOmaaqaaiabeg7aHnaaCaaaleqabaGaaGinaaaakiabgkHiTiab % eg7aHnaaCaaaleqabaGaaGOmaaaakiabgUcaRiabek7aInaaCaaale % qabaGaaGinaaaakiabgkHiTiabek7aInaaCaaaleqabaGaaGOmaaaa % kiabg2da9iaaicdaaeaacqGHshI3cqaHXoqydaahaaWcbeqaaiaais % daaaGccqGHRaWkcqaHYoGydaahaaWcbeqaaiaaisdaaaGccqGH9aqp % cqaHXoqydaahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGydaahaa % WcbeqaaiaaikdaaaaakeaacqGHshI3daqadaqaaiabeg7aHnaaCaaa % leqabaGaaGOmaaaakiabgUcaRiabek7aInaaCaaaleqabaGaaGOmaa % aaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaa % ikdadaqadaqaaiabeg7aHjabek7aIbGaayjkaiaawMcaamaaCaaale % qabaGaaGOmaaaakiabg2da9iabeg7aHnaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiabek7aInaaCaaaleqabaGaaGOmaaaakiaabccadaqada % qaaiaaboeacaqGVbGaaeyBaiaabchacaqGSbGaaeyzaiaabshacaqG % PbGaaeOBaiaabEgacaqGTaGaae4uaiaabghacaqG1bGaaeyyaiaabk % hacaqGLbaacaGLOaGaayzkaaaabaGaeyO0H49aaeWaaeaacqaHXoqy % daahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGydaahaaWcbeqaai % aaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH % 9aqpcqaHXoqydaahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGyda % ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaWaaeWaaeaacqaHXoqy % cqaHYoGyaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakeaacq % GHshI3daqadaqaaiaadggacqGHRaWkcaWGIbaacaGLOaGaayzkaaWa % aWbaaSqabeaacaaIYaaaaOGaeyypa0JaamyyaiabgUcaRiaadkgacq % GHRaWkcaaIYaGaamyyaiaadkgacqGHKjYOcaaIYaWaaeWaaeaacaWG % HbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqaba % GaaGOmaaaaaOGaayjkaiaawMcaaiaabccadaqadaqaaiaaboeacaqG % HbGaaeyDaiaabogacaqGObGaaeyEaiaab2cacaqGtbGaae4yaiaabI % gacaqG3bGaaeyyaiaabkhacaqG6bGaaeylaiaabMeacaqGUbGaaeyz % aiaabghacaqG1bGaaeyyaiaabYgacaqGPbGaaeiDaiaabMhaaiaawI % cacaGLPaaaaeaacqGHshI3caWGHbGaey4kaSIaamOyaiabgsMiJoaa % bmaabaGaamyyamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadkgada % ahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacqGHRaWkdaqadaqa % aiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWbaaS % qabeaacaaIYaaaaaGccaGLOaGaayzkaaGaeyOeI0IaaGOmaiaadgga % caWGIbaabaGaeyO0H4TaamyyaiabgUcaRiaadkgacqGHKjYOdaqada % qaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWba % aSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaey4kaSYaaeWaaeaaca % WGHbGaeyOeI0IaamOyaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaaaOqaaiabgkDiEpaabmaabaGaamyyaiabgkHiTiaadkgaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHLjYSdaqadaqaaiaa % dggacqGHRaWkcaWGIbaacaGLOaGaayzkaaGaeyOeI0YaaeWaaeaaca % WGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqa % baGaaGOmaaaaaOGaayjkaiaawMcaaiabgwMiZkaaicdaaeaacqGHsh % I3daqadaqaaiaadggacqGHRaWkcaWGIbaacaGLOaGaayzkaaGaeyyz % Im7aaeWaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaam % OyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaqaaiabgkDi % EpaabmaabaGaeqySde2aaWbaaSqabeaacaaIYaaaaOGaey4kaSIaeq % OSdi2aaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaeyyzIm7a % aeWaaeaacqaHXoqydaahaaWcbeqaaiaaisdaaaGccqGHRaWkcqaHYo % GydaahaaWcbeqaaiaaisdaaaaakiaawIcacaGLPaaaaeaacqGHshI3 % daqadaqaaiabeg7aHnaaCaaaleqabaGaaGOmaaaakiabgUcaRiabek % 7aInaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiabgkHiTmaa % bmaabaGaeqySde2aaWbaaSqabeaacaaI0aaaaOGaey4kaSIaeqOSdi % 2aaWbaaSqabeaacaaI0aaaaaGccaGLOaGaayzkaaGaeyyzImRaaGim % aaaaaa!4654! $$
Since the last equation includes 0, I cannot show any contradiction. I think the problem is that I did not use my hypothesis, ${\alpha ^2} + {\beta ^2}>2. % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaGaeqySde2aaWbaaSqabeaacaaIYaaaaO % Gaey4kaSIaeqOSdi2aaWbaaSqabeaacaaIYaaaaOGaaGOmaaaa!34DD! $
I think this claim can be proven using contradiction by I'm just not sure how to use my hypothesis.
An easier solution. Use Lagrange multipliers for $f(x,y) = x^2+y^2$ with the constraint $$g(x,y) = x^4 - x^2 + y^4 - y^2 = 0.$$ Putting $\nabla f(x,y)$ and $\nabla g(x,y)$ in rows: $$\begin{vmatrix} x & y \\ 2x^3 - x & 2y^3-y\end{vmatrix}=0 \implies xy^2-yx^3=0.$$ If $x \neq 0$ and $y \neq 0$ (the trivial case), we get $y = \pm x$. Back to $g(x,y) = 0$. $$x^4-x^2+x^2-x^2 = 0 \implies 2x^4-2x^2=0 \implies x^2-1 = 0 \implies x = \pm 1.$$ So we get the points $(1,1)$ and $(1,-1)$. And $f(1,1) = f(1,-1)= 2$. So the square of the maximum distance is $2$, hence the maximum distance is $\sqrt{2}$.