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let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$

How prove this inequality. Thanks

a suggestion please to prove this inequality

sharon
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  • Hello, a suggestion please to prove this inequality – sharon Mar 03 '15 at 19:14
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    Try to take the logarithm on both sides and rearrange the terms. – Clement C. Mar 03 '15 at 19:15
  • I feel I should point out that, if $a=b=c$, then you have equality - thus, the $<$ should be a $\leq$, since no further restraint is placed on $a$, $b$, and $c$ except "positive" and "real". – Glen O Mar 07 '15 at 04:29
  • This question has come up several times already, I think: http://math.stackexchange.com/questions/109783/prove-aabbcc-ge-abc-fracabc3-for-positive-numbers?lq=1 – Macavity Mar 09 '15 at 12:21

4 Answers4

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Let's transform $a^ab^bc^c\geq(abc)^{(a+b+c)/3}$ by taking $\log$: $$ a^ab^bc^c\geq(abc)^{(a+b+c)/3}\iff a\log a+b\log b+c\log c\geq\frac{a+b+c}{3}\log(abc)\\ \iff\frac{a}{a+b+c}\log a+\frac{b}{a+b+c}\log b+\frac{c}{a+b+c}\log c\geq\log(\sqrt[3]{abc}). $$ Let's prove the last inequality above. With $f(x)=-\log x=\log(1/x)$ being a convex function on $(0,\infty)$, we can use Jensen's inequality to infer that $$ \frac{a}{a+b+c}f(1/a)+\frac{b}{a+b+c}\log f(1/b)+\frac{c}{a+b+c}\log f(1/c)\\ \geq f\left(\frac{3}{a+b+c}\right)=\log\left(\frac{a+b+c}{3}\right)\geq \log(\sqrt[3]{abc}). $$ The last inequality above follows from the AM-GM inequality and the monotonicity of $\log$.

Kim Jong Un
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Without loss of generality suppose $a<b<c$. Then $\text{log} (a )<\text{log}( b) <\text{log} (c)$ and inequality of Chebyshev.

$\frac{a+b+c}{3}\frac{\text{log}(a)+\text{log}(b)+\text{log}(c)}{3}<\frac{a\text{log}(a)+b\text{log}(b)+c\text{log}(c)}{3}$.

Where you have to $a\text{log}(a)+b \text{log}(b)+c\text{log}(c)>\frac{a+b+c}{3}(\text{log}(a)+ \text{log}(b)+\text{log}(c)).$

And therefore $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$.

Angel
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    You cant award yourself your own bounty. Are you trying to draw attention for more up votes? You need at least 10 to break even now. – dustin Mar 06 '15 at 21:52
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    You get the inequality backwards in the last step :) (Also, I think you should allow the equality case with your signs, or say $a$, $b$, and $c$ are distinct). – Peter Woolfitt Mar 07 '15 at 21:01
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The inequality $a^ab^bc^c\le(abc)^{\frac{a+b+c}{3}}$ is incorrect. Let $a=b=1$, $c=4$

Then $$a^ab^bc^c=4^4$$ while $$(abc)^{\frac{a+b+c}{3}}=4^2$$.

However, the opposite inequality $a^ab^bc^c\ge(abc)^{\frac{a+b+c}{3}}$ is correct as demonstrated in Angel's answer.

Peter Woolfitt
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An elementary proof. No logs etc.

Notice that the inequality stays the same if we replace $a,b,c$ with $pa, pb, pc$:

$$ (pa)^{pa}(pb)^{pb}(pc)^{pc} = p^{pa+pb+pc} (a^ab^bc^c)^p$$

and

$$ (pa pb pc)^{(pa+pb+pc)/3} = p^{pa+pb+pc} ((abc)^{(a+b+c)/3})^p$$

Thus we can assume that $abc = 1$ (choose $p = \frac{1}{\sqrt[3]{abc}}$).

Now replace $c = 1/ab$ to try and prove

$$ \frac{a^a b^b}{(ab)^{\frac{1}{ab}}} \ge 1$$

Now there are two cases:

Case I)

$a \ge b \ge 1 \ge c$

In that case we easily have have $x = a- 1/ab \ge 0$ and so $a^a/a^{1/ab} = a^x \ge 1$ (and similarly with $b$).

Case II)

$a \le b \le 1 \le c$

In this case, we have that $x = a - 1/ab \le 0$, thus $a^x \ge 1$ (remember $a \le 1$). Similarly with $b$.

We are done!

Aryabhata
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