MT143统一分母

Posted 2020-10-31 mathstudy

已知$a,b>0$，则$m=\\dfrac{b^2+2}{a+b}+\\dfrac{a^2}{ab+1}$的最小值是______

解答:

$$m\\geqslant \\dfrac{b^2+2}{\\sqrt{a^2+1}\\cdot \\sqrt{1+b^2}}+\\dfrac{a^2}{\\sqrt{a^2+1}\\cdot \\sqrt{b^2+1}}=\\dfrac{(a^2+1)+(b^2+1)}{\\sqrt{a^2+1}\\cdot \\sqrt{b^2+1}}\\geqslant 2$$

当$a=1,b=1$时取到最小值$2$．