令z=b+a;则
由余弦定理得:
c^2=a^2 +b^2 -2·ab·cosC
=a^2 +b^2 -2·ab·cos30°
=a^2 +b^2 -√3·ab
=z^2 -(2+√3)ab;
即 z^2 -(2+√3)ab=(√6-√2)^2=8-4√3;
z=b+a≥2√(ab)→ab≤z^2/4;
则 z^2 -(2+√3)·(z^2/4)=[(2-√3)/4]·z^2≤8-4√3=4·(2-√3)
→z^2≤16;
z≤4;
即b+a的最大值是 4