# Homework Help: Should be an easy proof

1. Oct 1, 2008

### nicksauce

1. The problem statement, all variables and given/known data
Show that for all positive a,b
a+b <= a^2/b + b^2/a

3. The attempt at a solution
This simplifies to showing that
(a+b)(ab) <= a^3 + b^3
a^2b + b^2a <= a^3 + b^3

But I'm not really sure where to go from here... any hint would be appreciated.

2. Oct 2, 2008

### Phrak

It's Interesting that a+b = a^2/b + b^2/a when a=b.

So I might presume that a+b < a^2/b + b^2/a when a not equal b.

Try b = a + epsilon, and eliminate b.

Apply it to a^2b + b^2a <= a^3 + b^3.

3. Oct 2, 2008

How does this look?

As noted, if $$a = b$$ we have equality. Since the original expression is symmetric in the two variables, we can assume W.L.O.G. that $$a < b$$.

Then
\begin{align*} a+b & \le \frac{a^2}{b} + \frac{b^2}{a} \\ \intertext{if and only if} a^2b + ab^2 & \le a^3 + b^3 \\ \intertext{if and only if} ab^2 - b^3 & \le a^3 - a^2 b \\ b^2(a-b) & \le a^2 (a-b) \\ \intertext{if and only if} b^2 & \le a^2 \end{align*}

and, since the square function is strictly increasing on the positives, the final statement is true if and only if $$a < b$$.

Since equality holds when the two are equal, and inequality holds when they are not, we are done.

4. Oct 3, 2008

### Gib Z

Perhaps it would have been easier to start with nicksauce's idea:

Try factoring the RHS.

5. Oct 3, 2008