Bài 1:Cho $x,y,z\in (0,1)$.Chứng minh rằng: $(x-x^2)(y-y^2)(z-z^2)\ge (x-yz)(y-zx)(z-xy)$
Bài 2: Cho $x,y,z>0$ thỏa mãn: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$.Chứng minh rằng:
$\frac{x}{x^4+1+2xy}+\frac{y}{y^4+1+2yz}+\frac{z}{z^4+1+2zx}\le \frac{3}{4}$
Sau này lớn lên mn sẽ biết, biết nhiều ảnh hưởng đến sức khẻo –  tritanngo99 21-08-16 03:47 PM
chắc có hội này thật! search ông gu gồ mãi chả thấy đâu –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 21-08-16 03:42 PM
hư cấu =.= –  Confusion 21-08-16 03:41 PM
acamophomadady bodeshinobochocha hahahahaha –  tritanngo99 21-08-16 03:23 PM
tiếng thái lan! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 21-08-16 02:37 PM
omg, anh viết gì v? tiếng tàu ak? –  Confusion 21-08-16 02:36 PM
m?y th�m �i, ��y l� �? d?, b? �? th�i –  tritanngo99 21-08-16 02:34 PM
nghi ngờ cái hội này -_- –  Confusion 21-08-16 02:16 PM
bài 2 dễ quá! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 21-08-16 12:00 PM
2.
ta có $\frac{x}{x^{4}+1+2xy} \leq \frac{x}{2x^{2}+2xy}=\frac{1}{2}.\frac{1}{x+y}\leq \frac{1}{8}(\frac{1}{x}+\frac{1}{y})$
TT $\Rightarrow VT \leq \frac{1}{4}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{4}$
Dấu "=" $\Leftrightarrow x=y=z=1$
1.
giả sử $x\geq y\geq z$ khi đó do $x,y,z \in \left ( 0;1 \right )\Rightarrow  \begin{cases}x-yz>0 \\ y-zx>0 \end{cases}$
Nếu $z-xy<0$ BĐT lđ
Nếu $z-xy\geq 0$ khi đó ta cm $\sqrt{yz}(1-x)\geq \sqrt{(y-zx)(z-xy)} (1)$
Thật z $(1) \Leftrightarrow yz(1-x)^{2}\geq (y-zx)(z-xy)$
$\Leftrightarrow x(y-z)^{2}\geq 0$ (lđ)
TT ta suy ra đpcm 
Dấu "=" $\Leftrightarrow x=y=z$

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