What are the values of $B_1, B_2$, and $B_s$ so that the following inequality is satisfied with any value of $U_1$ and $U_2$? where $B_1, B_2, B_s, U_1$, and $U_2 \in \mathbb R^+$
$ B_1 U_1^2 + B_2 U_2^2 < B_s (U_1-U_2)^2 $
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First try $U_1=\cosh^2(x)\quad U_2=\sinh^2(x)$.Hope it helps to clarify something . – Miss and Mister cassoulet char May 29 '20 at 08:40
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Do you mean only $U_2\in\mathbb{R}^{+}$ ($B_1, B_2, B_s, U_1$ are real numbers)? – River Li Jun 08 '20 at 15:07
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Yes, all the numbers are positive reals – Jonattan Sarmiento Jun 08 '20 at 15:16
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What if $U_1 = U_2 > 0$ in your inequality? – River Li Jun 08 '20 at 15:30
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It seems that $B_1$ or $B_2$ must be negative. I will review the problem – Jonattan Sarmiento Jun 08 '20 at 16:56
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Hint: If $U_2 \neq 0$, consider the quadratic inequality
$$ (B_s - B_1) (\frac{U_1}{U_2} )^2 - 2 B_s (\frac{U_1}{U_2} ) + (B_s - B_2) > 0 $$
This has no real roots, so the necessary and sufficient condition is
1. $B_s - B_1 > 0 $,
2. $(2B_s)^2 - 4 (B_s - B_1)(B_s - B_2) < 0 $.
Can you proceed from here?
$ $
Deal with $U_2 = 0 $ separately
Conclude that the necessary and sufficient conditions are:
$$B_sB_1 + B_sB_2 - B_1B_2 < 0, B_s > B_1, B_s > B_2$$
Calvin Lin
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Thanks!! A single small detail: In the line $ (2B_s)^2 - 4 (B_s - B_1) ( B_s - B_2) < 0 $ Hence, $ B_s(B_1+B_2) - B_1 B_2 < 0 $ – Jonattan Sarmiento Jun 08 '20 at 07:55
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