Cho $a,b,c>0$. Tìm $\min P=\frac{4a^{3}+3b^{3}+2c^{3}-3b^{2}c}{(a+b+c)^{3}}$
Tìm $\min P=\frac{4a^{3}+3b^{3}+2c^{3}-3b^{2}c}{(a+b+c)^{3}}$
Started By duaconcuachua98, 06-11-2014 - 12:54
#1
Posted 06-11-2014 - 12:54
#2
Posted 07-11-2014 - 21:29
Cho $a,b,c>0$. Tìm $\min P=\frac{4a^{3}+3b^{3}+2c^{3}-3b^{2}c}{(a+b+c)^{3}}$
Ta có $P=\frac{4a^3+b^3+c^3+(2b^3+c^3-3b^2c)}{(a+b+c)^3}\geqslant \frac{4a^3+b^3+c^3}{(a+b+c)^3}$ theo AM-GM
Áp dụng BĐT Holder cho
$(4a^3+b^3+c^3)(\frac{1}{2}+1+1)(\frac{1}{2}+1+1) \geqslant (a+b+c)^3$
Vậy $P\geqslant \frac{4}{25}$
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