设a,b,c是直角三角形的三边,c为斜边,求最大常数M,使得1/a+1/b+1/c≥M/（a+b+c）

M≤3+(b/c+c/b)+(a/c+c/a)+(a/b+b/a) (b/c+c/b)*(a/c+c/a)=ab/c^2+b/a+a/b+c^2/ab=ab/(a^2+b^2)+b/a+a/b+(a^2+b^2)/(ab) =2(a/b+b/a)+1/(a/b+b/a) (b/c+c/b)^2+(a/c+c/a)^2=(b/c)^2+(c/b)^2+2+(a/c)^2+(c/a)^2+2 =4+b^2/c^2+1+a^2/b^2+a^2/c^2+1+b^2/a^2 =6+(b^2+a^2)/c^2+(a/b)^2+(b/a)^2=7+(a/b)^2+(b/a)^2 (b/c+c/b)+(a/c+c/a)=√[7+(a/b)^2+(b/a)^2+4(a/b+b/a)+2/(a/b+b/a)] M≤3+√[7+(a/b)^2+(b/a)^2+4(a/b+b/a)+2/(a/b+b/a)]+(a/b+b/a) 最大M可取到a/b=1时,即M=5+3√2