A + b + c = 1, tìm max e=can (2a ^ 2 + a + 1) +can(2b^ ^ 2+b+1) +căn(2c^2+c+1)

To find the maximum value of the expression, we can use the AM-GM inequality.

By AM-GM, we have:

(2a^2 + a + 1) + (2b^2 + b + 1) + (2c^2 + c + 1) ≥ 3√((2a^2 + a + 1)(2b^2 + b + 1)(2c^2 + c + 1))

Now, let's simplify the expression inside the cube root:

(2a^2 + a + 1)(2b^2 + b + 1)(2c^2 + c + 1) = (4a^4 + 2a^3 + 2a^2 + 2a + 2a^2 + a + 1)(2c^2 + c + 1)
= (4a^4 + 2a^3 + 4a^2 + 3a + 1)(2c^2 + c + 1)
= 8a^6 + 4a^5 + 8a^4 + 6a^3 + 2a^2 + 4a^4 + 2a^3 + 4a^2 + 3a + 1

Now, let's find the maximum value of this expression. Taking the derivative with respect to a and setting it equal to zero, we get:

d/dx (8a^6 + 4a^5 + 8a^4 + 6a^3 + 2a^2 + 4a^4 + 2a^3 + 4a^2 + 3a + 1) = 0
48a^5 + 20a^4 + 32a^3 + 18a^2 + 4a + 3 = 0

Unfortunately, this equation does not have a simple solution. However, we can use numerical methods or a graphing calculator to find the approximate value of a that satisfies this equation.

Once we have the value of a, we can substitute it back into the expression to find the maximum value.