1 reply to this topic

[DangHuyNgheAn]

Cho $-1\leq a,b,c\leq 1, a+b+c=0$. Chứng minh rằng $a^4+b^3+c^2\leq 2$.

Edited by leminhansp, 14-07-2014 - 10:20.
Latex

[canhhoang30011999]

Cho $-1\leq a,b,c\leq 1, a+b+c=0$. Chứng minh rằng $a^4+b^3+c^2\leq 2$.

ta có

$a^{2}(a-1)(a+1) \leq 0$

$ \Rightarrow a^{4}\leq a^{2}$

$b^{2}(b-1) \leq 0 \rightarrow b^{3} \leq b^{2}$

$ \Rightarrow a^{4}+b^{3}+c^{2}\leq a^{2}+b^{2}+c^{2}$

lại có $ (a-1)(b-1)(c-1) \leq 0$

$ (a+1)(b+1)(-c-1) \leq 0$

cộng vế vế ta có $-2ab-2bc-2ca-2 \leq 0$

$ \Rightarrow a^{2}+b^{2}+c^{2}\leq 2$

$ \Rightarrow a^{4}+b^{3}+c^{2}\leq 2$

Edited by canhhoang30011999, 14-07-2014 - 11:51.