一、選填題
G.
已知\(a,b\)為實數，試求\((3a-2b+1)^2+(2a+b-2)^2+(4a-5b-3)^2\)的最小值為
。
相關問題
https://math.pro/db/viewthread.php?tid=680&page=3#pid7957
H.
已知空間中一直線\(L\)：\(\displaystyle \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z}{1}\)與兩點\(A(0,1,3)\)、\(B(1,3,-2)\)。若點\(P\)為直線\(L\)上一動點，則\(\overline{PA}+\overline{PB}\)的最小值為
。
(我的教甄準備之路 兩根號的極值問題，
https://math.pro/db/viewthread.php?tid=661&page=3#pid22174)
P.
若方程組\(\cases{x(y+z-x)=39-2x^2 \cr y(x+z-y)=52-2y^2 \cr z(x+y-z)=78-2z^2}\)的正實數解為\(\cases{x=a\cr y=b\cr z=c}\)，則\(abc=\)
？
相關問題
https://math.pro/db/viewthread.php?tid=2020&page=1#pid11798
[解答]
\(\cases{xy+xz+x^2=39\cr yx+yz+y^2=52 \cr zx+zy+z^2=78}\)
三式相加\(2(xy+yz+zx)+x^2+y^2+z^2=169\)，\((x+y+z)^2=(\pm13)^2\)，\(x+y+z=\pm13\)
(1)當\(x+y+z=+13\)時
\(\cases{x(13-x)=39-x^2 \cr y(13-y)=52-y^2 \cr z(13-z)=78-z^2}\)，\(\cases{x=3\cr y=4\cr z=6}\)，\(xyz=72\)
(2)當\(x+y+z=-13\)時
\(\cases{x(-13-x)=39-x^2 \cr y(-13-y)=52-y^2 \cr z(-13-z)=78-z^2}\)，\(\cases{x=-3\cr y=-4\cr z=-6}\)，\(xyz=-72\)
二、計算證明題
1.
已知\(a,b,c\)為正實數，且滿足\(abc=1\)，試證明\(\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac{3}{2}\)。
Let \(a, b, c\) be positive real numbers such that \(abc = 1\). Prove that\(\displaystyle \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}\).
(1995IMO，
https://artofproblemsolving.com/ ... _Problems/Problem_2)
