# Proving a function is greater than zero (1 Viewer)

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#### Buri

The problem statement, all variables and given/known data

I'd like to show that:

f(x,y) = x^p/p + y^q/q - xy ≥ 0 where 1/q + 1/p = 1 and x ≥ 0 and y ≥ 0.

The attempt at a solution

What I'm actually trying to do is show that the minimum is on the line x = y^(q-1) at which the function takes on 0. But proving its actually always ≥ 0 is something that's causing me problems. Is there some inequality that could maybe lead me to a solution? I've tried considering lines in the plane like y = kx, but that doesn't lead me to anywhere. Any help?

#### micromass

We first take x fixed and we consider the function

$$h(y)=\frac{x^p}{p}+\frac{y^q}{q}-xy$$

The derivative of this function

$$h^\prime(y)=y^{q-1}-x$$

So the minimum of h is in in (q-1)th root of x. So all we need to prove is that

$$h(\sqrt[q-1]{x})=\frac{x^p}{p}+\frac{x^{\frac{q}{q-1}}}{q}-x^{\frac{q}{q-1}}=\frac{x^p}{p}-\frac{q-1}{q}x^{\frac{q}{q-1}}}$$

is positive. But since 1/p+1/q=1, we get that the above is equal to 0, and thus always positive.

It could be that I miscalculated somewhere, but I do think that this would be the technique to proving the inequality...

#### Buri

I've worked through it all and it doesn't seem like you did any miscalculations. Is it perfectly sound though? I don't know, but considering the single variable function is something I'm a bit iffy about (I'd think twice about it), but maybe it's just something clever to do.

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