Cho 2 số thực không âm  x;y  thỏa mãn x + y =1. Giá trị lớn nhất của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaamiEaaqaaiaadMhacqGHRaWkcaaIXaaaaiabgUca % RmaalaaabaGaamyEaaqaaiaadIhacqGHRaWkcaaIXaaaaaaa!4004! S = \frac{x}{{y + 1}} + \frac{y}{{x + 1}}\) là :

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaadMhacqGH9aqpcaaIXaGaeyi1HSTaamyEaiabg2da9iaaigda % cqGHsislcaWG4baaaa!4197! x + y = 1 \Leftrightarrow y = 1 - x\)

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaadaqadaqaaiaadIhaaiaawIcacaGLPaaaaeqaaOGaeyypa0Za % aSaaaeaacaWG4baabaGaaGymaiabgkHiTiaadIhacqGHRaWkcaaIXa % aaaiabgUcaRmaalaaabaGaaGymaiabgkHiTiaadIhaaeaacaWG4bGa % ey4kaSIaaGymaaaacqGH9aqpdaWcaaqaaiaadIhaaeaacaaIYaGaey % OeI0IaamiEaaaacqGHRaWkdaWcaaqaaiaaigdacqGHsislcaWG4baa % baGaamiEaiabgUcaRiaaigdaaaaaaa!50F8! {S_{\left( x \right)}} = \frac{x}{{1 - x + 1}} + \frac{{1 - x}}{{x + 1}} = \frac{x}{{2 - x}} + \frac{{1 - x}}{{x + 1}}\)  với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaaacaGLBbGaayzxaaaaaa!3C9B! x \in \left[ {0;1} \right]\).

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4uayaafa % Gaeyypa0ZaaSaaaeaacqGHsislcaaIXaaabaWaaeWaaeaacaaIYaGa % eyOeI0IaamiEaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaa % GccqGHRaWkdaWcaaqaaiabgkHiTiaaikdaaeaadaqadaqaaiaadIha % cqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaa % aakiabgYda8iaaicdaaaa!4827! S' = \frac{{ - 1}}{{{{\left( {2 - x} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} < 0\)  với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaaacaGLBbGaayzxaaaaaa!3C9B! x \in \left[ {0;1} \right]\)

Suy ra  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadg % gacaWG4bGaam4uaiabg2da9iaadofadaqadaqaaiaaicdaaiaawIca % caGLPaaacqGH9aqpcaaIXaaaaa!3F63! MaxS = S\left( 0 \right) = 1\)