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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.

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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 $S_0=3, S_{-1}=\frac 15$).

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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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