Linked Questions
20 questions linked to/from Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$
9
votes
6
answers
5k
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Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ [duplicate]
How can I prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$
I have tried substitute $x,y$ for numbers, which turns out right, but I don't know how to reason here.
Thanks in advance!
- 2,247
7
votes
3
answers
31k
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How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$? [duplicate]
I am reading a chapter about mathematical proofs. As an example there is:
Prove that: $$(1) \space\space\space\space\space\space\space\space\space\space\space \frac{a+b}{2} \geq \sqrt{ab}$$ for $a,b&...
- 571
1
vote
2
answers
5k
views
Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$ [duplicate]
Help I'm really stuck on a proof question:
Prove that, for all positive integers for $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$
Thanks!
- 51
4
votes
3
answers
705
views
How to go upon proving $\frac{x+y}2 \ge \sqrt{xy}$? [duplicate]
I'm trying to prove this but am having some difficulty.
For any $x,y\in\mathbb R$ such that $x\ge 0$ and $y\ge 0$ we have
$$\frac{x+y}2 \ge \sqrt{xy}.$$
So far what I have gotten to is $\frac{x+y}{...
- 149
2
votes
4
answers
366
views
If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$ [duplicate]
If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$
So far I've got:
$a<b$
$a^2<ba$
$a<\sqrt{ab}$
And:
$a<b$
$a+b<2b$
$\frac{a+b}{2}<b$
So I need to prove ...
- 1,435
0
votes
2
answers
880
views
Show that $\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}$ [duplicate]
Question:
Show that the harmonic mean is less than or equal to the arithmetic mean, and also less than or equal to the geometric mean, with equality if and only if $a=b$ ; ie., show that $$\frac{a+...
- 7,798
0
votes
5
answers
767
views
Show that if $x$ and $y$ are nonnegative reals, then $\frac{x + y}{ 2} ≥ \sqrt{xy}$ [duplicate]
Show that if $x$ and $y$ are nonnegative reals, then
$$\frac{x + y}{2} ≥ \sqrt{xy}$$
Can anybody help me?, I'm not quite sure how to prove this.
- 33
2
votes
3
answers
336
views
Prove that $\frac{a+b}{2}≥\sqrt{ab}$ [duplicate]
I'm working on a proof of AM-GM inequality, but for now I would like to prove more basic propositions.
Prove that the arithmetic-geometric mean inequality holds for lists of
numbers of length $2$...
- 3,371
1
vote
1
answer
335
views
Let $a>0$ and $b>0$. Prove that $\sqrt{ab} \le (a+b)/2$. [duplicate]
Let $a>0$ and $b>0$. Prove that $\sqrt{ab} \le (a+b)/2$.
Here is what I have tried:
Let $a \le b$. Multiplying both sides of this inequality by $a$ results in $a^2 \le ab$. It follows that $a \...
- 13
-2
votes
2
answers
97
views
Proofs my Contradiction [duplicate]
I have to solve the following problem by prooving using Contradiction. The problem is:
Show that for any nonnegative real numbers $a$ and $b$ we have: $\frac {a+b}2 ≥ \sqrt {ab}$
(Sadly, i dont even ...
- 41
0
votes
1
answer
36
views
Having trouble proving Inequality [duplicate]
I am having trouble proving this inequality: $2ab\leq a^2+b^2$
I can transpose the equation and change around signs. But I am not sure If I need to use k+1 here or just prove the inequality. In ...
- 231
17
votes
6
answers
7k
views
Arithmetic mean. Why does it work?
I've been using the formula for the arithmetic mean all my life, but I'm not sure why it works.
My current intuition is this one:
The arithmetic mean is a number that when multiplied by the number ...
- 1,760
2
votes
5
answers
10k
views
Prove $\sqrt{ xy} \leq \frac{x + y}{2}$ for all positive $x$ and $y$
First and foremost, I have already gone through the following posts:
Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$
Proving the AM-GM inequality for 2 numbers $\...
- 275
3
votes
4
answers
6k
views
Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$
I have to prove that
$$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$
The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I ...
- 9,324
2
votes
5
answers
305
views
Inequality Proof in Real Numbers: If $p+q=1$ then $pq\le\frac14$
Let $p,q \in \mathbb R$, and $p+q=1$. Prove that $$pq \le \frac{1}{4}.$$
The first thing I did was define 3 possibilities that we can have from $p+q=1$:
1º $p$ or $q$ is negative. Example:
$p=-10$...
- 3,312