KKTMinimize[obj_,eqns_,ineqns_,variables_]:=
Block[{myrule,lambda,u,\[Lambda],Lagrange,eqnu,ineqnlam,
eqnus,ineqnlams,kkteqns,kktvars,kktans},
myrule={z_[i_]:>ToExpression[ToString[z]<>ToString[i]]};
eqnu=u[#]&/@Range[Length@eqns];
ineqnlam=\[Lambda][#]&/@Range[Length@ineqns];
eqnus=If[Length@eqns>=1,eqns.eqnu,0];
ineqnlams=If[Length@ineqns>=1,ineqns.ineqnlam,0];
kktvars=Flatten@{variables,eqnu,ineqnlam}/.myrule;
Lagrange=obj-eqnus-ineqnlams;
kkteqns=Flatten@{Thread[D[Lagrange,{variables}]==0],
Thread[eqns==0],Thread[ineqns<=0],
Thread[ineqns*ineqnlam==0],Thread[ineqnlam<=0]}/.
myrule;
If[MemberQ[PolynomialQ[#,kktvars]&/@kkteqns[[All,1]],False],
Print["KKT限制式均需為多項式。"],
kktans=Reduce[kkteqns,kktvars,
Backsubstitution->True]/.{And->List,Or->List,
Equal->Rule};
If[Length@Dimensions@kktans==1,{obj/.kktans,
kktans},{obj/.#,#}&/@kktans]]];
Plot[(1-Cos[x]/Cos[x/2])/x^2,{x,0.-10,Pi/2},
PlotRange->{{0,Pi/2},{0.3,0.5}},Frame->True,
FrameTicks->{{{0.3,0.5,{3/8,"3/8"},{4/Pi^2,"4/Pi^2"}},
None},{{0,Pi/2},None}},
GridLines->{None,{3/8,4/Pi^2}},GridLinesStyle->Dashed]
D[(1-Cos[x]/Cos[x/2])/x^2,x]//Simplify
D[(1-(Cos[x/2]^2-Sin[x/2]^2)/Cos[x/2])/x^2,x]//Simplify
D[4-4Sec[x/2]+3xTan[x/2]-4Tan[x/2]^2+xTan[x/2]^3,x]//TrigExpand//Simplify
(D[(6x-2Sin[x/2]-4Sin[x]-2Sin[(3x)/2]+Sin[2x]),x]//TrigExpand)
/.{Sin[z_]:>a,Cos[z_]:>b}
KKTMinimize[6+4a^2+2a^4-b+9a^2b-4b^2-12a^2b^2-3b^3+2b^4,{},
{-a,-b,a-1/Sqrt[2],1/Sqrt[2]-b,b-1},{a,b}]
Limit[(1-(Cos[x/2]^2-Sin[x/2]^2)/Cos[x/2])/x^2,x->0]
Limit[(1-(Cos[x/2]^2-Sin[x/2]^2)/Cos[x/2])/x^2,x->Pi/2]
Let f(x)=(1-cos(x)/cos(x/2))/x^2. Taking the first derivative of f(x) with respect to x yields d/dx f(x)=cos(x/2)*g(x)/(2x^3),
where g(x)=x*tan^3(x/2)-4*tan^2(x/2)+3*x*tan(x/2)-4*sec(x/2)+4. Because g'(x)=1/4*sec(x/2)^2*h(x) and h'(x)=k(x)=2*sin^4(x/2)+4*sin^2(x/2)+2*cos^4(x/2)-3*cos^3(x/2)-4*cos^2(x/2)-cos(x/2)-12*sin^2(x/2) *cos^2(x/2)+9*sin^2(x/2) cos(x/2)+6. Let a=sin(x/2) and b=cos(x/2). By Karush-Kuhn-Tucker conditions, we know that k(x) reaches its minimum at a=0 and b=1. This, together with the facts that h(0)=0, g(0)=0, and f(0)=0, indicates that f(x) is increasing in x, which also implies that 3/8<(1-Cos[x]/Cos[x/2])/x^2<4 br="" completes="" proof.="" the="" this="">
