`-` Ta có:

`(1+a)(1+b) = 9/2 1+a+b+ab=9/2 a+b+ab=7/2`

Theo bất đẳng thức ` AM-GM` ta có:

`ab \leq ( (a+b)/2)^2`

`-> 7/2 - (a+b) \leq ((a+b)^2)/4`

`-> (a+b)^2 +4(a+b) -14 \geq 0`

`-> (a+b+2)^2 \geq 18`

`-> a+b+2 \geq \sqrt{18} =3\sqrt{2}`

`-> a+b \geq 3\sqrt{2} -2`

`(a+b)^2 + 4(a+b) -14 \geq 0`

`(a+b+2)^2 \geq 18 a+b \geq 3 \sqrt{2} -2 `

Ta có:

`a^2 +b^2 = (a+b)^2 -2ab = (a+b)^2 - 2(7/2 - a-b)= (a+b)^2 +2(a+b) -7= (a+b+2)^2 -11 \geq 18-11 =7`

Theo bđt Cauchy với 2 số dương (`a^4+1;b^4+1`) `:`

`- a^4 +1 \geq 2 \sqrt{a^4 xx1} = 2a^2`

`-> \sqrt{a^4+1} = (a^4 +1)/( \sqrt{a^4+1})`

` -b^4 +1 \geq 2 \sqrt{b^4 xx1} = 2b^2`

`->\sqrt{b^4+1} = (b^4 +1)/( \sqrt{b^4+1})`

Xét `Q^2= ( \sqrt{a^4+1} + \sqrt{b^4+1})^2 = a^4 +1 +b^4 +1 +2\sqrt{(a^4+1)(b^4+1)}`

`Q^2 = a^4 +b^4 + 2 +2\sqrt{(ab)^4 +a^4 +b^4+1}`

`-`Ta có: `a^4 +b^4= (a^2+b^2)^2 -2(ab)^2`

`-` Có: `2(ab)^2 \leq ((a^2+b^2)^2)/2`

`4(ab)^2 \leq (a^2+b^2)^2`

`4(ab)^2 \leq a^4 +2(ab)^2 +b^4`

`0\leq a^4+b^4 -2(ab)^2`

`0 \leq (a^2-b^2)^2`

Dấu `'='` xảy ra `<=> a^2=b^2`

`-> a=b`

`->(1+a)^2 =9/2`

`-> 1+a = \sqrt{9/2} = 3/\sqrt{2}` (Do `a+1>0`)

`-> a= 3/\sqrt{2} -1 =b`

Khi đó: `Q = 2xx \sqrt{(3/\sqrt{2} -1)^4 +1}`

`Q= 6\sqrt{2} -1`