Xét các số phức z=a+bia,b∈ℝ thỏa mãn z−4−3i=5. TínhP=a+b khi z+1−3i+z−1+i đạt giá trị lớn nhất.

Chọn A

Sử dụng BĐT Bunyakovsky

Từ giả thiết  $\left|z-4-3i\right|=\sqrt{5}$$\iff {\left(a-4\right)}^2+{\left(b-3\right)}^2=5$$\iff a^2+b^2-8a-6b+20=0$$\iff a^2+b^2=8a+6b-20$        

Mặt khác $T=\left|z+1-3i\right|+\left|z-1+i\right|$$=\sqrt{{\left(a+1\right)}^2+{\left(b-3\right)}^2}+\sqrt{{\left(a-1\right)}^2+{\left(b+1\right)}^2}$    

Suy ra $T^2\le \left(1^2+1^2\right)\left[{\left(a+1\right)}^2+{\left(b-3\right)}^2+{\left(a-1\right)}^2+{\left(b+1\right)}^2\right]$$=2\left[2\left(a^2+b^2\right)-4b+12\right]$$=2\left[2\left(8a+6b-20\right)-4b+12\right]$$=8\left(4a+2b-7\right)$           

Dấu = xảy ra khi ${\left(a+1\right)}^2+{\left(b-3\right)}^2={\left(a-1\right)}^2+{\left(b+1\right)}^2$$\iff a-2b=-2$    

Lại có  $4a+2b=4\left(a-4\right)+2\left(b-3\right)+22\le \sqrt{\left(4^2+2^2\right)\left[{\left(a-4\right)}^2+{\left(b-3\right)}^2\right]}+22$$=\sqrt{20.5}+22=32$     

Dấu = xảy ra khi $\frac{a-4}{4}=\frac{b-3}{2}\iff a-2b=-2$   

suy ra  $T^2\le 8\left(4a+2b-7\right)\le 8\left(32-7\right)=200$

 $\Rightarrow T\le 10\sqrt{2}$

Vậy $T_{\text{max}}=10\sqrt{2}$  khi  $\left\{\begin{array}{l}4a+2b=32\\ a-2b=-2\end{array}\right.$$\iff \left\{\begin{array}{l}a=6\\ b=4\end{array}\right.$  . Vậy  $a+b=10$.