<s>[INST] 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. 
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our conditions give $u=\frac{5}{3}$ and $w^3=15v^2$ and since
$$0\leq(a-b)^2(a-c)^2(b-c)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$
we obtain:
$$v^2(3v^2+25)^2\leq0$$ or $v^2\leq0$.
Thus, $$a^3+b^3+c^3=27u^3-27uv^2+3w^3=125+30v^2\leq125.$$
The equality does not occur because for this we need $abc=0$, which is impossible,
but $125$ is a supremum, of course. 
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