回復 1# newsonica 的帖子
非選第2題
已知非負實數\(a,b,c,d\)滿足\(a+2b+3c+4d=1\)且\((a+b)(c+d)^2=(a+c)(b+d)^2=k.\)則\(k\)的最大值為何?又該最大值在何時發生?
[解答]
\(\begin{align}
& \left( 1 \right)b=c \\
& \frac{1}{3}=\frac{a+2b+3c+4d}{3}=\frac{\left( a+b \right)+2\left( c+d \right)+2\left( c+d \right)}{3}=\frac{\left( a+c \right)+2\left( b+d \right)+2\left( b+d \right)}{3}\ge \sqrt[3]{4k} \\
& k\le \frac{1}{108} \\
\end{align}\)
等號成立於\(b=c=\frac{1}{3}-a=\frac{1}{6}-d\)
\(\begin{align}
& \left( 2 \right)b>c \\
& \frac{1}{3}=\frac{a+2b+3c+4d}{3}>\frac{\left( a+b \right)+2\left( c+d \right)+2\left( c+d \right)}{3}\ge \sqrt[3]{4k} \\
& k<\frac{1}{108} \\
& \\
& \left( 3 \right)b<c \\
& \frac{1}{3}=\frac{a+2b+3c+4d}{3}>\frac{\left( a+c \right)+2\left( b+d \right)+2\left( b+d \right)}{3}\ge \sqrt[3]{4k} \\
& k<\frac{1}{108} \\
\end{align}\)