将不等式k+2t≤k(a-b)+t(a+b)化简为：k + 2t ≤ ka - kb + ta + tb移项可得：2kb ≤ k + ta + tb合并同类项得：2k + 4t ≥ k + ta + tb即：k + 2t ≤ k(a-b) + t(a+b) ≤ 2k + 4t
将不等式k+2t<(k+t)a-(k-t)b<2k +4t化简为：k + 2t < ka + ta - kb + tb < 2k + 4t
移项可得：2kb < k + (2t-ta-tb) a < 2k + 4t - (ta + tb)因为-1 < cosθ < 1, 因此存在0 < θ < π/2，使得cosθ = (2t-ta-tb)/(2b), 然后将它代入上面的不等式中可得：k + 2t < (k + (2t-ta-tb) a)/(cosθ) < 2k + 4t - (ta + tb) / (cosθ)即：k + 2t < (k+t)a-(k-t)b < 2k +4t