<s>[INST] let $ \ \ 0let $ \ \ 0<a \leq b \leq c  \in \mathbb{R}$ 
then prove that :
$$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$

I do not know where to start please help me !
 [/INST] Your inequality is true!
We need to prove that
$$\frac{a+b+c}{3}-\frac{3abc}{ab+ac+bc}\geq\frac{(c-a)^2}{6c}$$ or
$$\frac{\sum\limits_{cyc}c(a-b)^2}{ab+ac+bc}\geq\frac{(c-a)^2}{2c}$$ or
$$2c(a-b)^2+2a(b-c)^2\geq\left(\frac{ab}{c}+a-b\right)(c-a)^2.$$
Now, by C-S
$$2c(a-b)^2+2a(b-c)^2=2ac\left(\frac{(b-a)^2}{a}+\frac{(c-b)^2}{c}\right)\geq$$
$$\geq\frac{2ac(b-a+c-b)^2}{a+c}=\frac{2ac(c-a)^2}{a+c}.$$
Thus, it remains to prove that
$$2ac\geq(a+c)\left(\frac{ab}{c}+a-b\right)$$ or
$$c^2a+c^2b\geq a^2b+a^2c,$$ which is obvious.
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