Chủ đề này có 2 trả lời

[hoctrocuaZel]

$(1):x,y,z>0.FindMin:A=\frac{3.\sum x^3}{4.\sum xy}+\frac{1}{(\sum x)^2}$

$(2):a,b,c>0:\prod a=\frac{9}{4}.Prove:\sum a^3> \sum a\sqrt{b+c}$

 

[Lee LOng]

(2).Ta có:

$a^{3}+b^{3}+c^{3}\geq 3abc=\frac{27}{4}$ $(*)$

$\sum a\sqrt{b+c}\leq \frac{(a^{2}+b^{2}+c^{2})+2(a+b+c)}{2}$

Ta chứng minh :$2\sum a^{3}\geq \sum a^{2}+2\sum a$

Ta có: $a^{3}+\frac{9}{4}+\frac{9}{4}\geq 3a\sqrt[3]{\frac{81}{16}}$

$\Rightarrow \frac{2\sum a^{3}+\frac{9}{4}.2.2.3}{3\sqrt[3]{\frac{81}{16}}}\geq 2\sum a$

Tương tự: $\Rightarrow \frac{2\sum a^{3}+\frac{9}{4}.3}{3\sqrt[3]{\frac{9}{4}}}\geq \sum a^{2}$

$Đpcm\Leftrightarrow (2-\frac{2}{3\sqrt[3]{\frac{81}{16}}}-\frac{2}{3\sqrt[3]{\frac{9}{4}}}).\sum a^{3}\geq \frac{9}{\sqrt[3]{\frac{81}{16}}}+\frac{9}{4\sqrt[3]{\frac{9}{4}}}$ (Đúng do $(*)$)

[Lee LOng]

(1). Ta có: 

$(\sum x^{3}).(\sum x)\geq (\sum x^{2})^{2}$

$\Rightarrow \frac{3.\sum x^{3}}{4\sum xy}\geq \frac{3(\sum x^{2})^{2}}{4(\sum xy)(\sum x)}\geq \frac{x+y+z}{4}$

$\Rightarrow A\geq \frac{x+y+z}{4}+\frac{1}{(x+y+z)^{2}}\geq 3\sqrt[3]{\frac{(x+y+z)^{2}}{64(x+y+z)^{2}}}=\frac{3}{4}$