题目不明确.是3√2/(2bc)还是(3√2/2)bc?是(3√2/2)bc设b=√(1-a²)sinβ,c=√(1-a²)cosβ,∵a,b,c＞0∴0<β＜π／2则,原式f(a,b,c)=a(b+c)+(3√2/2)bc=a*√(1-a²)(sinβ+cosβ)+(3√2/2)*(1-a²)sinβcosβ=a√[2(1-a²)]sin(β+π/4)+(3√2/4)*(1-a²)*sin(2β).............sin(2β)=-sin(-2β)=-cos(2β+π/2)=a√[2(1-a²)]sin(β+π/4)+(3√2/4)*(1-a²)*[-cos(2(β+π/4))]=a√[2(1-a²)]sin(β+π/4)+(3√2/4)*(1-a²)*[2sin²(β+π/4)-1]=a√[2(1-a²)]sin(β+π/4)+(3√2/2)*(1-a²)*sin²(β+π/4)-(3√2/4)*(1-a²)令√[2(1-a²)]sin(β+π/4)=x,√(1-a²)＜x≤√[2(1-a²)]则,原式f(a,b,c)=g(x)=ax+(3√2/4)x²-(3√2/4)*(1-a²)∵函数g(x)在√(1-a²)＜x≤√[2(1-a²)],a>0内单调递增,∴g|max=g(√[2(1-a²)])=a√[2(1-a²)]+(3√2/4)*2(1-a²)-(3√2/4)*(1-a²)=a√[2(1-a²)]+(3√2/4)*(1-a²),此时,x=√[2(1-a²)],sin(β+π/4)=1,∵0<β＜π／2,∴β=π/4,即b=c=√[(1-a²)/2],a²+2b²=1可设a=sinθ,b=cosθ/√2,0<θ＜π／2∴原式f(a,b,c)=a(b+c)+(3√2/2)bc=2ab+(3√2/2)b²=(2sinθcosθ)/√2+(3√2/4)cos²θ……cos²θ=(cos2θ+1)/2=(1/√2)sin2θ+(3√2/8)cos2θ+3√2/8……xsinα+ycosα=√(x²+y²)sin(α+ψ),其中ψ=arccos[x/√(x²+y²)]=[√(1/2﹢9/32)]*sin(2θ+ψ)+3√2/8……此时,ψ=arccos(4/5)≈36.9º=[√(25/32)]sin(2θ+ψ)+3√2/8=5√2/8sin(2θ+ψ)+3√2/8此时,原式最大值为5√2/8+3√2/8=√2,此时,θ=[π/2-arccos(4/5)]/2,a=√5/5,b=√10/5∴当a=√5/5,b=c=√10/5时,ab+ac+3√2/2bc取最大值,且最大值为√2...........希望对你有帮助!!!不懂,可以Hi我.....