# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Ex-ex-exponents! Topic: Inequalities. Level: AIME. August 20th, 2006

Problem: Let be positive reals. Prove that

.

Solution: How do you get rid of exponents? Well, take the log of course! Logging both sides and using log rules (multiplication to addition and bringing the power down), the inequality becomes

.

We will prove that

and

.

If we add those together, we get the inequality above. But we note that and are similarly sorted, we can apply the Rearrangement Inequality (see here), which tells us that

and

,

exactly what we needed. Thus the inequality is proven. QED.

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Comment: Classic way of proving inequalities with exponents; many times, after you take the log, you can apply Jensen’s (log is concave) to get some interesting results as well.

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Practice Problem: Prove AM-GM. For positive reals ,

.

### 7 Responses to “Ex-ex-exponents! Topic: Inequalities. Level: AIME.”

1. Moo Says:

are you kidding?
the practice problem is soooooooooooooooo easy! haha

Then do it =P.

3. Moo Says:

if you know Jensen’s inequality, the AM-GM inequality follows easily from the convexity of Lnx

but anyways here is a proof (dont know if the Latex will work but oh well):

Define

$x^m_k=\frac {a_m+a_{m+1}+\ldots a_{m+k}}{k}$

The inequality $x_1+x_2\geq 2\sqrt {x_1x_2}$ is the trivial inequality. Clearly if it holds for $n=2^t$, then

$x^1_{2^t}+x^{2^t+1}_{2^{t+1}}\geq 2\sqrt {x^1_{2^t}x^{2^t+1}_{2^{t+1}}}\geq 2^{t+1}\sqrt[2^t]{a_1a_2\ldots a_{2^{t+1}}}$ (ponder this for a minute). This establishes it for all powers of $2$ by induction.

Next, say that it holds for $n=k$. I claim that it also holds for $n=k-1$

Indeed, we have

$kx^1_{k-1}=a_1+a_2+\ldots a_{k-1}+x^1_{k-1}\geq k\sqrt[k]{a_1a_2\ldots a_{k-1}x^1_{k-1}}$, or

$x^1_{k-1}\geq \sqrt[k]{a_1a_2\ldots a_{k-1}x^1_{k-1}}$, which simplifies to

$(x^1_{k-1})^{\frac {k-1}{k}}\geq \sqrt[k]{a_1a_2\ldots a_{k-1}}$

which simplifies to $x^1_{k-1}\geq \sqrt[k-1]{a_1a_2\ldots a_{k-1}}$

which is just the AM-GM inequality for $n=k-1$

It’s easy to see that we can get it to work for any $n\in\mathbb{N}$ from here.

4. Moo Says:

guess not…..
for some of you Latex Masters it wouldn’t be hard to understand….haha

5. QC Says:

The practice problem is interesting. In fact, Jensen’s is equivalent to an interesting theorem about what I like to call function-means.

Define the function-mean of a function f and of some reals a_1, a_2, … a_n to be

f^{-1} ( ( f(a_1) + f(a_2) + … + f(a_n) ) / n }

For f a power function this gives power mean, for f a logarithm it gives geometric mean, and so forth.

Then Jensen’s is equivalent to the following statement:

Let M_{f} be the function-mean of f. Given two continuous, differentiable, one-to-one functions f, g and given some reals a_1, a_2, … a_n,

M_{f} \ge M_{g}

If and only if f(g^{-1}(x)) is convex. The reverse if it is concave.

Setting f = x, g = ln x produces AM-GM.

6. QC Says:

Alternately, it’s fairly obvious by Rearrangement.

7. Moo Says:

very true,
but first you have to prove Rearrangement…which is not hard to show.