Can Arithmetic Mean >= Geometric Mean be applied in this way

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Find the minimum value of the sum of real numbers a^(-5), a^(-4), 3*a^(-3), 1, a^8 and a^10 when a>0.

ignoring 1 for now, as it can be added later

We can find the solution by breaking 3*a^(-3) into a^(-3),a^(-3),a^(-3) and then applying Arithmetic Mean >= Geometric Mean for all positive real numbers.

But when I am breaking a^(-5) into 3 parts, a^(-4) into 3 parts, 3*a^(-3) into 3 parts, a^8 into 2 parts and a^10 into 2 parts, (such that both numerator and the denominator have a^36), I am not getting the same answer.

(1/3)*a^(-5)+(1/3)*a^(-5)+(1/3)*a^(-5) = P

(1/3)*a^(-4)+(1/3)*a^(-4)+(1/3)*a^(-4) = Q

a^(-3)+a^(-3)+a^(-3) = R

(1/2)*a^8+(1/2)*a^8 = S

(1/2)*a^10+(1/2)*a^10) = T

Now again applying Arithmetic Mean >= Geometric Mean

(P+Q+R+S+T)/13 >= (product of the terms)^(1/13)

Is this an appropriate logic or may be I did some silly mistake.