Đáp án+Giải thích các bước giải:
`1.13\sqrt{1/a^2+1/b^2+1/c^2}`
`=\sqrt{1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)-2/(ab)-2/(bc)-2/(ca)}`
`=\sqrt{(1/a+1/b+1/c)^2-(2(a+b+c))/(abc)}`
`=\sqrt{(1/a+1/b+1/c)^2}`
`=|1/a+1/b+1/c|`
`1.14)\sqrt{1+2013^2+2013^2/2014^2}+2013/2014`
`=\sqrt{2013^2+2.2013+1-2.2013+2013^2/2014^2}+2013/2014`
`=\sqrt{(2013+1)^2-2.2013+2013^2/2014^2}+2013/2014`
`=\sqrt{2014^2-2*2013/2014*2014+2013^2/2014^2}+2013/2014`
`=\sqrt{(2014-2013/2014)^2}+2013/2014`
`=|2014-2013/2014|+2013/2014`
`=2014-2013/2014+2013/2014=2014`
`1.15)x+y>=2\sqrt{xy}(x,y>=0)`
`<=>x-2\sqrt{xy}+y>=0`
`<=>(\sqrt{x}-\sqrt{y})^2>=0AAx,y>=0`
Dấu "=" xảy ra khi `x=y>=0`
`1/x+1/y+1/z>=1/\sqrt{xy}+1/\sqrt{yz}+1/\sqrt{zx}`
Áp dụng bất đẳng thức cosi ta có:
`1/x+1/y>=2/\sqrt{xy}`
`1/y+1/z>=2/\sqrt{yz}`
`1/z+1/x>=2/\sqrt{zx}`
`=>2(1/x+1/y+1/z)>=2(1/\sqrt{xy}+1/\sqrt{yz}+1/\sqrt{zx})`
`=>1/x+1/y+1/z>=1/\sqrt{xy}+1/\sqrt{yz}+1/\sqrt{zx}`(dpcm)
Dấu "=" xảy ra khi `x=y=z.`