x+y+z=1x + y + z = 1x+y+z=1 条件 g=x+y+z−1g = x + y + z - 1g=x+y+z−1 とおく H=−{xlogx+ylogy+zlogz}H = -\{x \log x + y \log y + z \log z\}H=−{xlogx+ylogy+zlogz} gx=gy=gz=1g_x = g_y = g_z = 1gx=gy=gz=1 Hx=−1log2(logx+1)Hy=−1log2(logy+1)Hz=−1log2(logz+1)H_x = -\frac{1}{\log 2}(\log x + 1) \qquad H_y = -\frac{1}{\log 2}(\log y + 1) \qquad H_z = -\frac{1}{\log 2}(\log z + 1)Hx=−log21(logx+1)Hy=−log21(logy+1)Hz=−log21(logz+1) ラグランジュの乗数法から {−1log2(logx+1)−λ=0−1log2(logy+1)−λ=0−1log2(logz+1)−λ=0\begin{cases} -\dfrac{1}{\log 2}(\log x + 1) - \lambda = 0 \\[2mm] -\dfrac{1}{\log 2}(\log y + 1) - \lambda = 0 \\[2mm] -\dfrac{1}{\log 2}(\log z + 1) - \lambda = 0 \end{cases}⎩⎨⎧−log21(logx+1)−λ=0−log21(logy+1)−λ=0−log21(logz+1)−λ=0 これから logx=logy=logz\log x = \log y = \log zlogx=logy=logz よって x=y=z=13x = y = z = \dfrac{1}{3}x=y=z=31 このとき H=log23H = \log_2 3H=log23