Maximum value of $a^3 + b^3 + c^3$ If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers.
My Attempt
writing $a+b=5-c$ and $ \frac{1}{a} +\frac{1}{b}  =\frac{1}{5} -\frac{1}{c}$ and after algebraic manipulation I obtain $(a+b)(\frac{1}{ab} + \frac{1}{5c})=0$. If $(a+b)=0$ we get $c=5$ and $a^3 + b^3 +c^3 =125$. But I can't analyze the case $(\frac{1}{ab} + \frac{1}{5c})=0$. Can anyone help me from this point. Thanks in advance. 

Here is a lesser known approach to these things, which sometimes helps
Let $xy+yz+zx=p$ so that $x^2+y^2+z^2=(x+y+z)^2-2p=25-2p$.
Set $S_3=a^3+b^3+c^3$
Then clearing fractions in the second equation gives $abc=5p$.
We have also that $a,b,c$ are roots of the polynomial equation $$q(x)=x^3-5x^2+px-5p=0$$
Now observe that $0=q(a)+q(b)+q(c)=S_3-5(a^2+b^2+c^2)+p(a+b+c)-15p$ so that $$S_3-125+10p+5p-15p=0$$ and $S_3=125$

Observe also that we have, with $$S_r=a^r+b^r+c^r$$$$a^{n-3}q(a)+b^{n-3}q(b)+c^{n-3}q(c)=0=S_n-5S_{n-1}+pS_{n-2}-5pS_{n-3}$$which gives us the sums of powers in terms of $p$ using the recurrence relation (note that $S_0=3, S_{-1}=\frac 15$).
