切线法

对于${\displaystyle x_{1},x_{2},...,x_{n}\in D,x_{1}+x_{2}+...+x_{n}=k}$ ，D为给定区间，k为常数，求证${\displaystyle f(x_{1})+f(x_{2})+...+f(x_{n})\leq (\geq )C}$ 

观察得取等条件为${\displaystyle x_{1}=x_{2}=...=x_{n}={\frac {k}{n}}}$ 时，找出${\displaystyle f(x)}$ 在${\displaystyle x={\frac {k}{n}}}$ 处的切线函数${\displaystyle px+q}$ （假设f可导），尝试证明局部不等式${\displaystyle f(x)\leq (\geq )px+q}$ 。^[2]

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

猜得${\displaystyle a=b=c={\frac {1}{3}}}$ 时取等，构造切线使${\displaystyle {\sqrt {4a+1}}\leq pa+q}$ 让${\displaystyle {\sqrt {4a+1}}+{\sqrt {4b+1}}+{\sqrt {4c+1}}\leq {\sqrt {21}}}$ 成立。

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

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

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

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

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