Using the fact that $$cd\leq\frac{c^p}{p}+\frac{d^q}{q}$$ if $$\frac{1}{p}+\frac{1}{q}=1$$ and letting $$c=\frac{f(t)}{\left(\int_{a}^{b}f(t)^p dt\right)^\frac{1}{p}}$$ and $$d=\frac{g(t)}{\left(\int_{a}^{b}g(t)^q dt\right)^\frac{1}{q}},$$ I have proven a lemma which states $$\int_{a}^{b}f(t)g(t)dt\leq\left(\int_{a}^{b}f(t)^p dt\right)^\frac{1}{p}\left(\int_{a}^{b}g(t)^q dt\right)^\frac{1}{q}.$$ But, how can this be used to prove the triangle inequality for this norm? I am a little confused about the $p$'s and $q$'s. Do I use what I know about the relationship of $p$ and $q$ to write the $q$'s as $p$'s?
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What happens if you apply your fact using the values of $c$ and $d$ you have, and integrate the resulting inequality from $a$ to $b$? – Aaron Jan 26 '12 at 04:47

perhaps [this](http://math.stackexchange.com/q/87636/8271) can interest you – leo Jan 26 '12 at 07:14

2I would post this as a comment if I could... This inequality is called Minkowski's inequality. You can find a proof using Holder's inequality (your lemma) on the wikipedia page for Minkowski's inequality. – zzz Jan 26 '12 at 05:38
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HINT:
From the lemma you proved (Hölder's Inequality). Let $f,g \in L_{p}[a,b]$.
Then $$\int_{a}^{b}f+g^{p}=\int_{a}^{b}f+g^{p1}f+g$$
$$\le\int_{a}^{b}f+g^{p1}(f+g) \text{ by triangle inequality of absolute value function}$$
$$=\int_{a}^{b}f+g^{p1}f+\int_{a}^{b}f+g^{p1}g$$
From this step you can apply Hölder's inequality (for integrals), simplify by using the relationship of $p$ and $q$, then you will get your inequality called Minikwoski's Inequality.
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