I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$\frac{a+b+c}{3}\cdot \frac{a+b+c}{3}\geq \frac{ca+b^2+ca}{3}$$
My problem
I would've wanted to get $\dfrac{ab+bc+ca}{3}$ on the right instead. Did I apply the inequality wrong, or does it follow that $\dfrac{ab+bc+ca}{3}$ is less than or equal to $\dfrac{a+b+c}{3}\cdot \dfrac{a+b+c}{3}$?
Best Answer
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain: $$3(a\cdot a+b\cdot b+c\cdot c)\geq(a+b+c)(a+b+c)$$or $$3(a^2+b^2+c^2)\geq(a+b+c)^2$$ or $$3(a^2+b^2+c^2+2(ab+ac+bc))\geq(a+b+c)^2+2(ab+ac+bc)$$ or $$(a+b+c)^2\geq3(ab+ac+bc)$$ or $$\frac{a+b+c}{3}\geq\sqrt{\frac{ab+ac+bc}{3}}.$$ By the same way we obtain: $$3(ab\cdot ab+ac\cdot ac+bc\cdot bc)\geq (ab+ac+bc)(ab+ac+bc)$$ or $$3(a^2b^2+a^2c^2+b^2c^2)\geq(ab+ac+bc)^2$$ or $$(ab+ac+bc)^2\geq3abc(a+b+c).$$ Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain: $$(ab+ac+bc)^3\geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)\geq$$ $$\geq3abc\cdot3(c\cdot ab+b\cdot ac+a\cdot bc)=27a^2b^2c^2,$$ which gives $$\sqrt{\frac{ab+ac+bc}{3}}\geq\sqrt[3]{abc}.$$