<s>[INST] Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
  $$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$

I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.
 [/INST] Ok, here we go.

Let $f(a,b,c,\lambda)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a^3+b^3+c^3}{3}+\lambda(a+b+c-3)$.

$\nabla f=0$ means:


*

*$\frac{\partial f}{\partial a}=-a^2-\frac{1}{a^2}+\lambda=0$

*$\frac{\partial f}{\partial b}=-b^2-\frac{1}{b^2}+\lambda=0$

*$\frac{\partial f}{\partial c}=-c^2-\frac{1}{c^2}+\lambda=0$
Now,
$ \frac{\partial f}{\partial b}- \frac{\partial f}{\partial a}=0 <=> a^2-b^2+\frac{1}{a^2}-\frac{1}{b^2}=0 <=>$

$(a-b)(a+b)+(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}+\frac{1}{b})=0 <=>$

$(a-b)[(a+b)-\frac{1}{ab}(\frac{1}{a}+\frac{1}{b})]=0 <=>$

$(a-b)(a+b)(1-\frac{1}{(ab)^2})=0$

And of course the same for all other combinations.

So $a=b$ or $ab=1$. Also, $b=c$ or $bc=1$.

Suppose $a=b$. $b=c$ doesn't work (we can test to see it's not a minimum) so we must have $bc=1$.

Hence we search a value of $a$ with $2a+\frac{1}{a}=3$. This is $a=\frac{1}{2}$. We can test to see this is the minimum (with a value of $\frac{7}{4}$), and we're done.

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