若a,b,c>0，且ax2+bx+c=0有實根，試證明：max{a,b,c}>=(4/9)(a+b+c)

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The art of doing mathematics consists in finding that special case which contains all the germs of generality

(4/9)(a+b+c)>max{a,b,c}
設max{a,b,c}=b
則因為4(a+c)>5b
4c>4(a-b)+4c>b
0>(b-4c)(b-c)
5bc-4c^2>b^2
但b^2>=4ac
so 5bc-4c^2>4ac
=> 5b>4(a+c) (矛盾)
設設max{a,b,c}=/=b
則令a>=b
b>=b^2/4a>4c
因此5b>=4(a+c)>5a => b>a (矛盾)
所以max{a,b,c}>=(4/9)(a+b+c)

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想要得到什麼,就必須付出相同的代價...這就是煉金術中所說的"等價交換原則".

ax2+bx+c=0有實根 => b^2-4ac≧0

(1)a=max{a,b,c}時
a^2≧b^2≧4ac => a≧4c => a/4≧c
a+a+(a/4)≧a+b+c => a≧(4/9)(a+b+c)

(2)c=max{a,b,c}時，同(1)之論證方法

(3)b=max{a,b,c}時
b-(4/9)(a+b+c)
=(1/9)(5b-4a-4c)
=(1/9b)[5b^2-4b(a+c)]
=(1/9b)[(b^2-4ac)+4b^2-4b(a+c)+4ac]
=(1/9b)[(b^2-4ac)+4(b-a)(b-c)]
≧0

由(1)(2)(3)得證max{a,b,c}≧(4/9)(a+b+c)

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The art of doing mathematics consists in finding that special case which contains all the germs of generality