08/07/17 22:10:13
私にも解けますた…
>>387
(1) a/(1+a) + b/(1+b) ≧ a/(1+a+b) + b/(1+a+b) = (a+b)/(1+a+b) ≧ c/(1+c). {← x/(1+x) は単調増加}
∵ (a+b)(1+c) - (1+a+b)c = (a+b) - c ≧ 0.
(2) a+b=s, b-a=d とおくと
(左辺) = {a+(1/a)}^2 + {b+(1/b)}^2
= (a^2 + b^2) + {(1/a)^2 + (1/b)^2} + 4
= (a^2 + b^2){1 + (1/ab)^2} + 4
= (1/2)(s^2 + d^2){1 + 16/(s^2 - d^2)^2} + 4
これは d^2 について単調増加。d=0 のとき最小値
(1/2)(s^2){1 + (2/s)^4} + 4 = 2{(s/2)+(2/s)}^2.
(別法) f(x) = {x + (1/x)}^2 = x^2 + 2 + 1/(x^2) は下に凸だから
f(a) + f(b) ≧ 2f((a+b)/2) = 2f(s/2) = 2{(s/2)+(2/s)}^2.
(3) 基本対称式を x+y+z =s, xy+yz+zx =t, xyz =u とおく。
(xy+yz+zx)/(xyz) = t/u ≧ 3*(3/s),
(x+y+z)/(xyz) = s/u ≧ 3*(3/s)^2,
1/(xyz) = 1/u ≧ (3/s)^3,
(左辺) = {2+(1/x)}{2+(1/y)}{2+(1/z)} = {8xyz + 4(xy+yz+zx) + 2(x+y+z) + 1}/(xyz)
= 8 + 4(xy+yz+zx)/(xyz) + 2(x+y+z)/(xyz) + 1/(xyz)
≧8 + 12*(3/s) + 6*(3/s)^2 + (3/s)^3
= {2 + (3/s)}^3
>>388
(4) 相乗相加平均より
(与式) ≦ {1 - (x+y+z)/(3n)}^3 = {1 - s/(3n)}^3.