[B.3989] a,b,c are positive real numbers, such that a^2 +b^2 +c^2 +abc = 4. Prove that a+b+c ≦ 3. (略解) 1-a,1-b,1-c のうち2つは同符号、よって (1-b)(1-c) ≧0 としてもよい。 3(3-a-b-c) + (a^2 +b^2 +c^2 +abc -4) = (1/2)(2-a-b)^2 + (1/2)(2-b-c)^2 + (1/2)(2-c-a)^2 -(1-a)(1-b)(1-c) = (1/4)(3-2c-a)^2 + (1/4)(3-2b-a)^2 + (1/2)(1-a)^2 + a(1-b)(1-c) ≧ a(1-b)(1-c).
[C.892] Prove that if x,y,z are positive real numbers and xyz=1, the values of the expressions 1/(1+x+xy), y/(1+y+yz), xz/(1+z+xz) cannot all be greater than 1/3. (略解) 1/(1+x+xy) = yz/(1+y+yz) = z/(1+z+xz), xy/(1+x+xy) = y/(1+y+yz) = 1/(1+z+xz), x/(1+x+xy) = 1/(1+y+yz) = xz/(1+z+xz), 辺々たす.