7. [tex]\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy} \geq \frac{4}{x^2+y^2+2xy}+\frac{1}{\frac{(x+y)^2}{2}}=\frac{4}{(x+y)^2}+\frac{2}{(x+y)^2}=\frac{6}{(x+y)^2}=6[/tex]
8. [tex]\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y})+\frac{x+y}{2xy}+\frac{2}{x+y} \geq \frac{1}{\sqrt{xy}}+2\sqrt{\frac{x+y}{2xy}.\frac{2}{x+y}}=1+2\frac{1}{\sqrt{xy}}=1+2=3[/tex]
9. [tex]x^4+y^4 \geq \frac{1}{2}(x^2+y^2)^2 \geq \frac{1}{2}[\frac{1}{2}(x+y)^2]^2=\frac{1}{8}(x+y)^4=2[/tex]
10. [tex]x^3+y^3=(x+y)(x^2-xy+y^2) \geq 2(x^2-xy+y^2)=(x^2+y^2)+(x-y)^2 \geq x^2+y^2[/tex]
11. [tex]\frac{1}{a+1}+\frac{1}{b+1} \geq \frac{4}{a+b+2}=\frac{4}{3}[/tex]
12. 1)[tex]\frac{yz}{x}+\frac{zx}{y} \geq 2\sqrt{\frac{yz}{x}.\frac{zx}{y}}=2z[/tex]
Tương tự rồi cộng vế theo vế, chia 2 cả 2 ta có đpcm.
2) [tex]\frac{x}{y^2}+\frac{1}{x} \geq 2\sqrt{\frac{x}{y^2}.\frac{1}{x} }=\frac{2}{y}[/tex]
Tương tự cộng vế theo vế, rút gọn ta có đpcm.
13. [tex]\frac{x^3}{y}+xy \geq 2\sqrt{\frac{x^3}{y}.xy}=2x^2[/tex]
Cộng vế theo vế tương tự ta có: [tex]\frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x}+xy+yz+zx\geq 2(x^2+y^2+z^2)\geq 2(xy+yz+zx)\Rightarrow \frac{x^3}{y}+\frac{y^3}{z}\geq xy+yz+zx[/tex]
BĐT phụ được sử dụng: [tex]\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y}[/tex]
[tex]xy \leq \frac{(x+y)^2}{4}[/tex]
[tex]x^2+y^2 \geq \frac{1}{2}(x+y)^2[/tex]
[tex]x^2+y^2+z^2 \geq xy+yz+zx[/tex]