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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 enter image description here

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 $\dfrac{a+b+c}{3}\cdot \dfrac{a+b+c}{3}$?

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2 Answers 2

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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 $$\sqrt{\frac{ab+ac+bc}{3}}\geq\sqrt[3]{abc}.$$

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As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.

Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 \ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 \ge ab + bc + ca$.

The last equality is true, as we have that $a^2 + b^2 \ge 2ab$, $b^2 + c^2 \ge 2bc$ and $c^2 + a^2 \ge 2ac$. Add the inequalities to get the final answer.

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