`E = (x + y - 2,5)^(2026) + (y^2 + (1)/(3))^2 - 10`

Vì `{((x + y - 2,5)^(2026) >= 0 \quad \forall x, y), (y^2 >= 0 => y^2 + (1)/(3) >= (1)/(3) => (y^2 + (1)/(3))^2 >= (1)/(9) \quad \forall y):}`

`-> (x + y - 2,5)^(2026) + (y^2 + (1)/(3))^2 - 10 >= 0 + (1)/(9) - 10 \quad \forall x, y`

`-> E >= -(89)/(9)`

Dấu "=" xảy ra khi và chỉ khi:

$\left\{\begin{aligned} &x + y - 2,5 = 0 \\ &y = 0 \end{aligned}\right.$

$\left\{\begin{aligned} &x = 2,5 \\ &y = 0 \end{aligned}\right.$

Vậy `E_(\min) = -(89)/(9)` khi `x = 2,5` và `y = 0`

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