Tìm Min  $\left(x^4+1\right)\left(y^4+1\right)$$x+y=\sqrt{10}$với ; x, y > 0.

  $A=\left(x^4+1\right)\left(y^4+1\right)$ $=x^4{y}^4+1+x^4+y^4$

$\iff A={\left(x^2+y^2\right)}^2+{\left(x^2{y}^2-1\right)}^2$

$\iff A={\left[{\left(x+y\right)}^2-2xy\right]}^2+{\left(x^2{y}^2-1\right)}^2$

$\iff A={\left(10-2xy\right)}^2+{\left(x^2{y}^2-1\right)}^2$

Đặt t = xy $\Rightarrow A={\left(10-2t\right)}^2+{\left(t^2-1\right)}^2$

$\iff A=t^4+2t^2-40t+101$

Theo BĐT AM– GM: $xy\le {\left(\frac{x+y}{2}\right)}^2=\frac{5}{2}$  do đó $t\in \left(0;\frac{5}{2}\right]$

Thấy $A=t^4+2t^2-40t+101={\left(t^2-4\right)}^2+10{\left(t-2\right)}^2+45$

$\iff A={\left(t-2\right)}^2\left[{\left(t+2\right)}^2+10\right]+45\ge 45$

Dấu “=” xảy ra khi t = 2 (thỏa mãn điều kiện)

Vậy A min = 45 $\iff t=2\iff \left(x,y\right)=\left(\frac{\sqrt{10}-\sqrt{2}}{2},\frac{\sqrt{10}+\sqrt{2}}{2}\right)$  .