(这里x的平方=x^2)
证明:
首先证明 a^2/(a+b) + b^2/(b+c) + c^2/(c+a) = b^2/(a+b) + c^2/(b+c) + a^2/(c+a)
[a^2/(a+b) + b^2/(b+c) + c^2/(c+a)] - [b^2/(a+b) + c^2/(b+c) + a^2/(c+a)]
= [a^2/(a+b) - b^2/(a+b)] + [b^2/(b+c) - c^2/(b+c)] + [c^2/(c+a) - a^2/(c+a)]
= (a^2 - b^2)/(a+b) + (b^2 - c^2)/(b+c) + (c^2 - a^2)/(c+a)
= (a-b) + (b-c) + (c-a)
= 0
所以 a^2/(a+b) + b^2/(b+c) + c^2/(c+a) = b^2/(a+b) + c^2/(b+c) + a^2/(c+a).
然后证明 [a^2/(a+b) + b^2/(b+c) + c^2/(c+a)] + [b^2/(a+b) + c^2/(b+c) + a^2/(c+a)] ≥(a+b+c)/2
[a^2/(a+b) + b^2/(b+c) + c^2/(c+a)] + [b^2/(a+b) + c^2/(b+c) + a^2/(c+a)]
= [a^2/(a+b) + b^2/(a+b)] + [b^2/(b+c) + c^2/(b+c)] + [c^2/(c+a) + a^2/(c+a)]
= (a^2 + b^2)/(a+b) + (b^2 + c^2)/(b+c) + (c^2 + a^2)/(c+a)
因为 (a^2 + b^2) - [(a+b)^2]/2 = 1/2 * (a-b)^2 ≥ 0
所以 (a^2 + b^2) ≥ [(a+b)^2]/2,即 (a^2 + b^2)/(a+b) ≥ (a+b)/2
所以(a^2 + b^2)/(a+b) + (b^2 + c^2)/(b+c) + (c^2 + a^2)/(c+a) ≥ (a+b)/2 +(b+c)/2 + (c+a)/2
以上不等式右边 = a+b+c
于是我们证得了(a^2 + b^2)/(a+b) + (b^2 + c^2)/(b+c) + (c^2 + a^2)/(c+a) ≥ a+b+c ---(1);
又因为 a^2/(a+b) + b^2/(b+c) + c^2/(c+a) = b^2/(a+b) + c^2/(b+c) + a^2/(c+a),
于是(1)两边都除以2,证得
a^2/(a+b) + b^2/(b+c) + c^2/(c+a) ≥ (a+b+c)/2
--------------------我是证明结束的分割线-------------------------
以下是吐槽:
好想用手写,打符号好麻烦 T_T