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Prove that

$$a^ab^bc^c\ge(abc)^{(a+b+c)/3}$$

where $a,b,c\in\mathbb{R^+}$

I tried using powered AM-GM but didn't get anything. please give me a hint to solve it.

Satvik Mashkaria
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4 Answers4

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Since $\log(x)$ is monotonically increasing, we have $$ (a-b)(\log(a)-\log(b))\ge0\\ (c-a)(\log(c)-\log(a))\ge0\\ (b-c)(\log(b)-\log(c))\ge0\\ $$ Add these and $$ (a\log(a)+b\log(b)+c\log(c))-(a\log(a)+b\log(b)+c\log(c))\ge0 $$ to get $$ 3(a\log(a)+b\log(b)+c\log(c))-(a+b+c)(\log(a)+\log(b)+\log(c))\ge0 $$ Divide by $3$ and exponentiate to get $$ a^ab^bc^c\ge(abc)^{\frac{a+b+c}3} $$

robjohn
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$$a^ab^bc^c\ge{(abc)^\frac{a+b+c}{3}} \stackrel{\log (\cdot)}{\iff} a\log a+b\log b+c\log c \geq (\log a+\log b+\log c)\cdot \frac{a+b+c}{3}$$

You can use Chebyshev's sum inequality to finish it.

UserX
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Take the $\log$ of both sides. This is equivalent to: $$ a\log a + b\log b + c\log c \ge \frac{a+b+c}3 (\log a + \log b + \log c) $$

now use the rearrangement inequality twice to get:

$$ a\log a + b\log b + c\log c \ge b\log a + c\log b + a\log c ;\\ a\log a + b\log b + c\log c \ge c\log a + a\log b + b\log c;\\ a\log a + b\log b + c\log c = a\log a + b\log b + c\log c. $$ The sum of these inequalities is the right result.

Alternative:

$$ \frac{a+b+c}3 \frac{\log a + \log b + \log c}3 \le \frac{a+b+c}3 \log \frac{a+b+c}3 $$using the concavity of $\log$; $$ \le \frac{a\log a + b\log b + c\log c }3 $$using convexity of $x\to x\log x$.

mookid
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First Prove that $$a^ab^b\ge a^bb^a$$ Then w have $$a^ab^b\ge a^bb^a$$ $$a^ac^c\ge a^cc^a$$ $$c^cb^b\ge b^cc^b$$

$$a^ab^bc^c\ge a^a b^bc^c$$

Multiplying together

$$a^{3a}b^{3b}c^{3c}\ge (abc)^{a+b+c}$$

$$a^{a}b^{b}c^{c}\ge (abc)^{(a+b+c)/3}$$

  • I did not downvote, but perhaps a good hint would be something like $\left(\frac ab\right)^{a-b}\ge1$ by considering the cases $a\ge b$ and $a\lt b$, – robjohn Oct 25 '14 at 13:08