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Suppose $A$, $B$, $C$ and $D$ are fixed arbitrary positive numbers.

I am free to choose $\epsilon$, $\epsilon_1$ and $\epsilon_2$ but they must be positive.

Can I choose the epsilons so that $$\left((A -\frac{1}{\epsilon_1} - \frac{B}{2\epsilon_2})\frac{C\epsilon}{(1+\epsilon)} - \epsilon_1 D\right) > 0?$$

I need this to show coercivity of a bilinear form.. Thanks.

I did some calculations, and it seems if I make $\epsilon_1$ very small and $\frac{\epsilon}{1+\epsilon}$ small then it might work but I can't prove it. Anyway I really hope it can be done.

Thanks

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    Have you tried using calculus to minimize the left hand side as a function of $\epsilon, \epsilon_1, \epsilon_2$? (Well, it's clear that $\epsilon_2 = +\infty$ is right; then if you can satisfy the inequality, you can later replace $\epsilon_2$ with a sufficiently large real number)2012-12-24
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    @Hurkyl I didn't try that.. I wanted to avoid the differentiation. But I will give it a go.2012-12-24

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Let $\alpha=\frac{1}{\epsilon_1}$, $\beta=\frac{1}{2\epsilon_2}$ and $\gamma=\frac{\epsilon}{1+\epsilon}$. Then you want to find a triple $(\alpha,\beta,\gamma)$ with $\alpha,\beta>0$ and $0<\gamma<1$ and, after dividing both sides by $\epsilon_1$ and rearranging, the condition:

$$\alpha(A-\alpha-B\beta)C\gamma > D$$

In particular, you need:

$$\alpha(A-\alpha-B\beta)> D/C$$

and if you do have this, you can easily find $\gamma$.

Now, the supremum of $\alpha(A-\alpha-B\beta)$ is the supremum of $\alpha(A-\alpha)$, which is when $\alpha=A/2$, and yields the value $\frac{A^2}4$. So at minimum, you need:

$$\frac{A^2}{4}> D/C$$

You can pick $(\alpha,\beta,\gamma)$ (and hence $\epsilon,\epsilon_1,\epsilon_2$) if and ony if this is true.

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    That's a shame, thanks.2012-12-24