Chủ đề này có 1 trả lời

[zack]

cho a,b,c >0, abc=1 cm

$\frac{a^{3}+3}{a^{3}(b+c)}+\frac{c^{3}+3}{b^{3}(a+c)}+\frac{b^{3}+3}{b^{3}(a+c)}\geq 6$

 

[25 minutes]

cho a,b,c >0, abc=1 cm

$\frac{a^{3}+3}{a^{3}(b+c)}+\frac{c^{3}+3}{b^{3}(a+c)}+\frac{b^{3}+3}{b^{3}(a+c)}\geq 6$

Ta có $\frac{a^3+3}{a^3(b+c)}\geqslant \frac{3a+1}{a^3(b+c)}=\frac{3}{a^2(b+c)}+\frac{1}{a^3(b+c)}$

$\Rightarrow\sum \frac{a^3+3}{a^3(b+c)}\geqslant \sum \frac{3}{a^2(b+c)}+\sum \frac{1}{a^3(b+c)}=A_1+A_2$

Xét $A_1=\sum \frac{3}{a^2(b+c)}$

Đặt $(a,b,c)=(\frac{1}{x},\frac{1}{y},\frac{1}{z})\Rightarrow xyz=1$

$\Rightarrow A_1=\sum \frac{3x^2}{\frac{1}{y}+\frac{1}{z}}=\sum \frac{3x^2yz}{y+z}=\sum \frac{3x}{y+z}\geqslant \frac{9}{2}$

Xét $A_2=\sum \frac{1}{a^3(b+c)}=\sum \frac{x^3}{\frac{1}{y}+\frac{1}{z}}=\sum \frac{x^3yz}{y+z}=\sum \frac{x^2}{y+z}\geqslant \frac{x+y+z}{2}\geqslant \frac{3}{2}$

Từ đó $\Rightarrow P\geqslant \frac{9}{2}+\frac{3}{2}=6$

Đẳng thức xảy ra khi $a=b=c=1$