Showing posts with label inequality. Show all posts
Showing posts with label inequality. Show all posts
Hard Inequality Problem: Prove that $6<3^{\sqrt{3}}<7$
Prove that $6<3^{\sqrt{3}}<7$ from the simple and straightforward inequality where $1<\sqrt{3}<2$.
This is a particularly daunting mathematics challenge and you could spend days or even a week trying to prove the inequality, with no fruitful result.
Notice that if we exponentiate the given inequality $1<\sqrt{3}<2$ with base 3, we get:
$3^1<3^{\sqrt{3}}<3^2$
$3<3^{\sqrt{3}}<9$ (Compare it with the targeted inequality $6<3^{\sqrt{3}}<7$)
The lower and upper bound that we could get from the given inequality is far too low and high for the targeted inequality.
This is a particularly daunting mathematics challenge and you could spend days or even a week trying to prove the inequality, with no fruitful result.
Notice that if we exponentiate the given inequality $1<\sqrt{3}<2$ with base 3, we get:
$3^1<3^{\sqrt{3}}<3^2$
$3<3^{\sqrt{3}}<9$ (Compare it with the targeted inequality $6<3^{\sqrt{3}}<7$)
The lower and upper bound that we could get from the given inequality is far too low and high for the targeted inequality.
Floor Function Problem...
Solve the following equation:
$\displaystyle \left\lfloor x+\frac{7}{3} \right\rfloor^2-\left\lfloor x-\frac{9}{4} \right\rfloor=16$
Note: $\displaystyle \lfloor x \rfloor$ denotes the largest integer not greater than $x$. This function, referred to as the floor function, is also called the greatest integer function, and its value at $x$ is called the integral part or integer part of $x$.
$\displaystyle \left\lfloor x+\frac{7}{3} \right\rfloor^2-\left\lfloor x-\frac{9}{4} \right\rfloor=16$
Note: $\displaystyle \lfloor x \rfloor$ denotes the largest integer not greater than $x$. This function, referred to as the floor function, is also called the greatest integer function, and its value at $x$ is called the integral part or integer part of $x$.
Trigonometric Inequality
Show that :
[math]\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2[/math]
[math]\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2[/math]
Prove that $x^2+y^2<1$.
Let $x,\,y>0$ be such that $x^3+y^3<x-y$. Prove that $x^2+y^2≤1$.
We have to rate this as a 5 star intriguing Mathematical Olympiad Inequality problem because it is a special problem that is designed for the application of heuristic skills.
I am going to show some sublimely insightful approaches of intelligent people here and as always, I hope you will to learn something awesome today!
We have to rate this as a 5 star intriguing Mathematical Olympiad Inequality problem because it is a special problem that is designed for the application of heuristic skills.
I am going to show some sublimely insightful approaches of intelligent people here and as always, I hope you will to learn something awesome today!
Olympiad Trigonometric Problem (Continued)
Previously we asked you, the educators or yourself, if you are a student to think of way(s) to find the ratio of $\dfrac{PR}{QR}$ in this thread (Olympiad Trigonometric Problem), we will now lead you to a credible way to solve for that intriguing hard trigonometric problem.
We in fact, should have the eagle eye sight and should be able to tell offhand that we suspect $\angle P=\angle Q$.
We in fact, should have the eagle eye sight and should be able to tell offhand that we suspect $\angle P=\angle Q$.
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