切線法

切線法是利用切線構造不等式的方法，有時會結合延森不等式。^[1]

切線法是屬於試探性的方法，但使用範圍比延森不等式更廣，例如半凹半凸的函數${\displaystyle x^{2}+2{\sqrt {x}}}$不能使用延森不等式，但能使用切線法。^[2]

常規方法

對於${\displaystyle x_{1},x_{2},...,x_{n}\in D,x_{1}+x_{2}+...+x_{n}=k}$，D為給定區間，k為常數，求證${\displaystyle f(x_{1})+f(x_{2})+...+f(x_{n})\leq (\geq )C}$

觀察得取等條件為${\displaystyle x_{1}=x_{2}=...=x_{n}={\frac {k}{n}}}$時，找出${\displaystyle f(x)}$在${\displaystyle x={\frac {k}{n}}}$處的切線函數${\displaystyle px+q}$（假設f可導），嘗試證明局部不等式${\displaystyle f(x)\leq (\geq )px+q}$。^[2]

例子

已知${\displaystyle a,b,c\geq 0,a+b+c=1}$，求證${\displaystyle {\sqrt {4a+1}}+{\sqrt {4b+1}}+{\sqrt {4c+1}}\leq {\sqrt {21}}}$

猜得${\displaystyle a=b=c={\frac {1}{3}}}$時取等，構造切線使${\displaystyle {\sqrt {4a+1}}\leq pa+q}$讓${\displaystyle {\sqrt {4a+1}}+{\sqrt {4b+1}}+{\sqrt {4c+1}}\leq {\sqrt {21}}}$成立。

${\displaystyle {\sqrt {4a+1}}={\sqrt {\frac {7}{3}}}}$

${\displaystyle ({\sqrt {4a+1}})'={\frac {2}{\sqrt {4a+1}}}=2{\sqrt {\frac {3}{7}}}}$

所求切線為${\displaystyle 2{\sqrt {\frac {3}{7}}}(a-{\frac {1}{3}})+{\sqrt {\frac {7}{3}}}}$

${\displaystyle (2{\sqrt {\frac {3}{7}}}(a-{\frac {1}{3}})+{\sqrt {\frac {7}{3}}})^{2}-(4a+1)={\frac {4}{21}}(3a-1)^{2}\geq 0}$證得${\displaystyle {\sqrt {4a+1}}\leq 2{\sqrt {\frac {3}{7}}}(a-{\frac {1}{3}})+{\sqrt {\frac {7}{3}}}}$

${\displaystyle {\sqrt {4a+1}}+{\sqrt {4b+1}}+{\sqrt {4c+1}}\leq 2{\sqrt {\frac {3}{7}}}(a+b+c-1)+3{\sqrt {\frac {7}{3}}}={\sqrt {21}}}$^[2]