難難難的雙參問題
已知函式\(f(x)=\ln(1+x)+\dfrac{x^2}{2}\)
(1)當\(x>0\)，比較\(f(x)\)與\(x\)的大小
(2)若函式\(g(x)=\cos x+\dfrac{x^2}{2}\)，且\(f\left(e^{\frac{a}{2}}\right)=g(b)-1(a>0,b>0)\)，證明:\(f(b^2)+1>g(a+1)\)
解
(1)做差，記\(\varphi(x)=f(x)-x=\ln(1+x)+\dfrac{x^2}{2}-x\)
\(\varphi^{\prime}(x)=\dfrac{1}{x+1}+x-1=\dfrac{1}{x+1}+x+1-2\geq 2-2=0\)
從而\(\varphi(x)\)單調遞增，則\(\varphi(x)\geq \varphi(0)=0\)
從而\(f(x)\geq x\)
(2)設\(h(x)=f(x)+1-g(x)=\ln(1+x)+1-\cos x\)，
當\(x>0\)時，\(1-\cos x\geq 0,\ln(1+x)>0\)，則\(h(x)>0\)恆成立
由\(h\left(e^{\frac{a}{2}}\right)>0\)，得\(f\left(e^{\frac{a}{2}}\right)+1>g\left(e^{\frac{a}{2}}\right)\)
從而\(g(b)>g\left(e^{\frac{a}{2}}\right)\)
不難證明\(g(x)\)單調遞增
則\(g(b)>g\left(e^{\frac{a}{2}}\right)\)得到\(b>e^{\frac{a}{2}}\)
同理\(h(b)=f(b)+1-g(b)>0\)
得\(f(b^2)+1>g(b^2)\)
則要證\(f(b^2)+1>g(a+1)\)
即證:\(g(b^2)>g(a+1)\)
即證：\(b^2>a+1\)
因\(b>e^{\frac{a}{2}}\)，則\(b^2>e^a\)
因\(e^a>a+1\)
則\(b^2>e^a>a+1\)