Is the Inequality a+b <= a^2/b + b^2/a Easy to Prove?

[nicksauce]

Homework Statement

Show that for all positive a,b
a+b <= a^2/b + b^2/a

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.

 

[Phrak]

Homework Statement

Show that for all positive a,b
a+b <= a^2/b + b^2/a

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.

 

[statdad]

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

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.

 

[Gib Z]

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

This simplifies to showing that
(a+b)(ab) <= a^3 + b^3

Try factoring the RHS.

 

[statdad]

Easier? Nay, infinitely easier (well, not infinitely, but much easier).
That's the problem with proofing one's own work - you get wrapped up and can't see the trees for the forest. Good point Gib Z.