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[chardhdmovies]

cho $a,b,c\in \left [ 1,2 \right ]$.CMR $a^3+b^3+c^3\leq 5abc$

 

NTP

[dogsteven]

Cách cho cấp 2:

Giả sử $a\geqslant b\geqslant c$, khi đó:

$$f(a,b,c)=a^3+b^3+c^3-5abc$$

$$f(a,b,c)-f(a,b,1)=(c-1)(c^2+c+1-5ab) \leqslant (c-1)(c+1-4ab) \leqslant 0$$

$$f(a,b,1)-f(a,1,1)=(b-1)(b^2+b+1-5a) \leqslant (b-1)(b^2-4b+1) \leqslant 0$$

$$f(a,1,1)=a^3-5a+2 =(a-2)(a^2+2a-1) \leqslant 0$$

Cách cho cấp 3:

Ta thấy $f(a,b,c)=a^3+b^3+c^3-5abc$ lồi nên $f(a,b,c) \leqslant \text{max}\{f(1,1,2), f(1,2,1), f(2,2,1), f(1,1,1), f(2,2,2)\}$

Edited by dogsteven, 15-12-2014 - 11:23.