Given real numbers $c \ge b \ge a>0$, prove that

$$ \frac{(c-a)^{2}}{6c} \le \frac{a+b+c}{3} - \frac{3}{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

*using well-known inequality

Other solution without famous inequality: I already know a solution that uses only algebraic manipulation here https://youtu.be/A0Qv8H1YDsc

But I am trying a solution that uses AM and HM inequality, because the RHS seems familiar to AM and HM.

Attempt: Notice obviously that $$\frac{a+b+c}{3} \ge (abc)^{1/3} $$ $$ - \frac{3}{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge -(abc)^{1/3} $$

But using these only imply the RHS of the problem is non-negative. While the left hand side can be positive depending on $a$,$c$.

Next, by write the 2nd term of RHS of the problem as

$$ \frac{3}{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} = \frac{3abc}{ab +bc +ac}$$

the inequality to be proven is

$$ \frac{(c-a)^{2}}{2} (a + b + \frac{ab}{c}) \le (a+b+c)(ab+ac+bc)-9abc$$

By AM-GM the RHS is bigger than or equal to: $$ 9(abc)^{1/3} (abc)^{2/3} - 9abc =0$$ too. So can we solved this problem using well-known inequality?



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