cho x,y,z>0 và $\sqrt[]{x}$ + $\sqrt[]{y}$ + $\sqrt[]{z}$ =1 tìm GTLN của: P= $\sqrt[]{xyz}$ ( $\frac{1}{\sqrt[]{(x+y)(y+z)} }$ + $\frac{1}{\sqrt[]{(x

Áp dụng BĐT CauChy ta có : 

`P =\sqrt{xyz}(1/(\sqrt{(x + y)(y + z)}) + 1/(\sqrt{(x + z)(y + z)}) + 1/(\sqrt{(x + y)(x + z)}))`

`<=> P = 1/2 \sqrt{xyz}(2/(\sqrt{(x + y)(y + z)}) + 2/(\sqrt{(x + z)(y + z)}) + 2/(\sqrt{(x + y)(x + z)})) `

`<= 1/2 \sqrt{xyz}(1/(x + y) + 1/(y + z) + 1/(x + z) + 1/(y + z) + 1/(x + y) + 1/(x + z)) `

`= \sqrt{xyz}(1/(x + y) + 1/(y + z) + 1/(z + x))`

`<= \sqrt{xyz}(1/(2\sqrt{xy}) + 1/(2\sqrt{yz}) + 1/(2\sqrt{zx}) )`

`= \sqrt{xyz}/(2\sqrt{xy}) + \sqrt{xyz}/(2\sqrt{yz}) + \sqrt{xyz}/(2\sqrt{zx}) `

`= (\sqrt{x} + \sqrt{y} + \sqrt{z})/2`

`= 1/2 (sqrt{x} + sqrt{y} + sqrt{z} =1 ) `

`Dấu “=” <=> x = y = z = 1/9`

Vậy Max `P =1/2`