Y = 4x ^ 2 + 16 / (x ^ 2 + 1) minimum value of function

Y = 4x ^ 2 + 16 / (x ^ 2 + 1) minimum value of function

y=4x^2+4-4+16/(x^2+1)
=4(x^2+1)+16/(x^2+1)-4
x^2+1>=1
4 (x ^ 2 + 1) + 16 / (x ^ 2 + 1) > = 2 radical [4 (x ^ 2 + 1) * 16 / (x ^ 2 + 1)] = 16
When 4 (x ^ 2 + 1) = 16 / (x ^ 2 + 1), take the equal sign
(x^2+1)^2=4
x^2=1
The equal sign can pass through
So Y > = 16-4
So the minimum value is 12

Minimum value of function y = 4x ^ 2 + 16 / (x ^ 2 + 1) ^ 2

Let x ^ 2 + 1 = t, then t > = 1
y=4(t-1) + 16/t^2 = 2t + 2t + 16/t^2 -4 >= 3(2t*2t*16/t^2)^1/3 - 4 = 3*8-4=20
When 2T = 16 / T ^ 2, that is, t = 2, the minimum value of 20 is obtained
Here x ^ 2 = 1

Find the minimum value of the function y = 4x ^ 2 + 16 / (2 + x ^ 2), and find the value of X when the minimum value is reached

Let u = x ^ 2 + 2
Then y = 4U + 16 / U - 8 (U > = 2)
From the basic inequality,
y >= 2 * sqrt(4u * 16/u) - 8 = 8
If and only if 4U = 16 / u, the equal sign holds
Therefore, combined with the condition of U > = 2, we can see that when u = 2, y takes the minimum value of 8
Here x = 0