Cho hai số thực x , y thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadIhadaahaaadbeqaaiaaikdaaaWccqGH % RaWkcaWG5bWaaWbaaWqabeaacaaIYaaaaaWcbeaakmaabmaabaGaam % iEaiabgUcaRiaadMhaaiaawIcacaGLPaaacqGHLjYScaaIXaaaaa!44A8! {\log _{{x^2} + {y^2}}}\left( {x + y} \right) \ge 1\). Tìm giá trị lớn nhất của biểu thức S = x+2y.

Điều kiện bài toán tương đương

Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikda % aaGccqGH8aapcaaIXaaaaa!3C76! {x^2} + {y^2} < 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iEaiabgUcaRiaadMhacqGHKjYOcaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaakiabgYda8iaaig % daaaa!4365! \Rightarrow x + y \le {x^2} + {y^2} < 1(1)\)

Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikda % aaGccqGH+aGpcaaIXaaaaa!3C7A! {x^2} + {y^2} > 1\) thì  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaadMhacqGHLjYScaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4k % aSIaamyEamaaCaaaleqabaGaaGOmaaaaaaa!3F50! x + y \ge {x^2} + {y^2}(2)\)

Với (1) ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizIm6aaO % aaaeaadaqadaqaaiaaigdadaahaaWcbeqaaiaaikdaaaGccqGHRaWk % caaIYaWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaWaaeWaae % aacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEamaaCaaa % leqabaGaaGOmaaaaaOGaayjkaiaawMcaaaWcbeaakiabgYda8maaka % aabaGaaGynaaWcbeaaaaa!45C0! S = x + 2y \le \sqrt {\left( {{1^2} + {2^2}} \right)\left( {{x^2} + {y^2}} \right)} < \sqrt 5 \)

Với (2) ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikda % aaGccqGHsislcaWG4bGaeyOeI0IaamyEaiabgsMiJkaaicdaaaa!40FB! {x^2} + {y^2} - x - y \le 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaaaiaa % wIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaai % aadMhacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaaGaayjkaiaa % wMcaamaaCaaaleqabaGaaGOmaaaakiabgsMiJoaalaaabaGaaGymaa % qaaiaaikdaaaaaaa!4849! \Leftrightarrow {\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} \le \frac{1}{2}\)

Khi đó sử dụng bất đẳng thức ta có:

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaamiEaiabgkHiTmaalaaabaGaaGymaaqaaiaaikda % aaaacaGLOaGaayzkaaGaey4kaSIaaGOmamaabmaabaGaamyEaiabgk % HiTmaalaaabaGaaGymaaqaaiaaikdaaaaacaGLOaGaayzkaaGaey4k % aSYaaSaaaeaacaaIZaaabaGaaGOmaaaaaaa!45D0! S = \left( {x - \frac{1}{2}} \right) + 2\left( {y - \frac{1}{2}} \right) + \frac{3}{2}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizIm6aaO % aaaeaadaqadaqaaiaaigdadaahaaWcbeqaaiaaikdaaaGccqGHRaWk % caaIYaWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaWaaeWaae % aadaqadaqaaiaadIhacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaa % aaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabm % aabaGaamyEaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaacaGL % OaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaale % qaaOGaey4kaSYaaSaaaeaacaaIZaaabaGaaGOmaaaaaaa!4E47! \le \sqrt {\left( {{1^2} + {2^2}} \right)\left( {{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {y - \frac{1}{2}} \right)}^2}} \right)} + \frac{3}{2}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizIm6aaO % aaaeaadaWcaaqaaiaaiwdaaeaacaaIYaaaaaWcbeaakiabgUcaRmaa % laaabaGaaG4maaqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaaIZaGaey % 4kaSYaaOaaaeaacaaIXaGaaGimaaWcbeaaaOqaaiaaikdaaaaaaa!40CF! \le \sqrt {\frac{5}{2}} + \frac{3}{2} = \frac{{3 + \sqrt {10} }}{2}\)

So sánh hai trường hợp suy ra giá trị lớn nhất của  là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaey4kaSYaaOaaaeaacaaIXaGaaGimaaWcbeaaaOqaaiaaikda % aaaaaa!39F9! \frac{{3 + \sqrt {10} }}{2}\)