Questions tagged [a.m.-g.m.-inequality]
For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.
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Show that : $\sqrt{[x]\cdot \{x\}} +\sqrt{x \cdot \{x\}} + \sqrt{[x]\cdot x} \leq 2x$
Show that for any positive real number $x$ the inequality holds:
$\sqrt{[x]\cdot \{x\}} +\sqrt{x \cdot \{x\}} + \sqrt{[x]\cdot x} \leq 2x$
where by $[a], \{a\}$ we mean the whole par and fractional ...
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Maximize $f(x)=(1-x)^5(1+x)(1+2x)^2$
For which value of $x$ is the product $(1-x)^5(1+x)(1+2x)^2$ a maximum, and what is this value?
This is easy with calculus, but how would you do it without calculus? $f(x)=(1-x)^5(1+x)(1+2x)^2 \geq 0$ ...
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Proving inequality where $a^2 +b^2 +c^2 =3$ and $a,b,c$ are positive reals
For $a,b,c >0 $ and given $a^2 +b^2 +c^2 =3 $ prove the following
inequality:
$\large a+b+c +\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \leqslant
3\left( \frac{a}{b} + \frac{b}{c} +\frac{c}{a} -1 \...
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Proving inequality given one constraint
Given $a,b,c>0$ and $a+b+c=1$ then show that
$$2+\frac{ab}{c} +\frac{bc}{a} +\frac{ca}{b} \leqslant \frac{1}{9abc}.$$
My work: the inequality is equivalent to
$$18abc + 9((ab)^2 +(bc)^2 +(ca)^2 ) \...
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Proving an inequality given a constraint
Given $a,b,c >0 $ and $ab+bc+ca=3$ prove the following inequality :
$\large 3\left( \frac1{a} + \frac1{b} +\frac1{c} \right) \geqslant 6 + \frac{ab}{c} +\frac{bc}{a} + \frac{ca}{b}$
My work : LHS$= ...
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I want help knowing if my solution is correct
If $a, b, x, y$ are positive rational numbers such that $\frac 1x + \frac 1y = 1$ then prove that $\frac {a^{x}}{x}+ \frac {b^{y}}{y}$ $\ge ab$
This question is from Problems Plus in IIT Mathematics. ...
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Given $A, B, C, D$ in $Oxyz$ space, find $M \in CD$ such that $MA + MB$ is smallest. Why can't I use AM-GM to solve this?
In the $Oxyz$ space, consider four points $A(-1, 1, 6),$
$B(-3,-2,-4),$ $C(1,2,-1),$ $D(2,-2,0).$ Find $M \in CD$ such that $△MAB$ has the smallest perimeter.
As $AB$ is constant, the task is ...
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Minimizing $\left(\frac{c}{a} + \frac{c}{b}\right)^2$, where $c$ is the hypotenuse of a right triangle with legs $a$ and $b$
This question is regarding the following problem
Given that $a, b, c$ are the sides of the $\triangle ABC$ which is right angled at $C$, then what is the minimum value of the following expression?
$$\...
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Prove $(x_1 +x_2 +...+x_i +...+x_n)^2 \ge n(x_1x_2 +x_2x_3 +...+x_ix_{i+1} +...+x_nx_1)$
Determine all positive integers $n \ge 2$ such that for all POSITIVE real number $x_1, x_2,..., x_n$ the following inequality holds:
$$(x_1 +x_2 +...+x_i +...+x_n)^2 \ge n(x_1x_2 +x_2x_3 +...+x_ix_{i+...
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prove that $\left( \sum_{i=1}^{n} a_i \right) \left( \sum_{i=1}^{n} a_i^{n-1} \right) \leq n \prod_{i=1}^{n} a_i + (n-1) \sum_{i=1}^{n} a_i^n.$
question:Let $a_1, a_2, \ldots, a_n$ be nonnegative real numbers. Prove that
$$\left( \sum_{i=1}^{n} a_i \right) \left( \sum_{i=1}^{n} a_i^{n-1} \right) \leq n \prod_{i=1}^{n} a_i + (n-1) \sum_{i=1}^{...
user1327314
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Knowing $x,y,z\ge0$ prove $x^2+xy^2+xyz^2\ge4xyz-4$
Knowing $x,y,z\ge0$ prove $x^2+xy^2+xyz^2\ge4xyz-4$
I thought that I should rearrange this inequality to be somewhat of the form of Schur's Inequality and WLOG I assumed $x\ge y\ge z$.
Trying this way ...
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Knowing $a,b,c>0$ and $abc\le1$, prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1+\frac{6}{a+b+c}$
Knowing $a,b,c>0$ and $abc\le1$, prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1+\frac{6}{a+b+c}$
I tried to AM-GM the inequality, which gave this:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\...
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A symmetric inequality involving product of three variables
Let $a, b,c \ge 0, ab + bc + ca + abc = 4$. Find the minimum of $S = \sqrt{a}+\sqrt{b}+\sqrt{c}$.
My guess is that $S$ attends its minimum at $b = c = 2, a = 0$ and the other permutation of $(2, 2, 0)$...
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Hard Inequalities between harmonic and geometric means
Let's have $0 < \lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n$ $n$ positive numbers
How to show that :
$$\left( \frac{n}{\sum_{k = 1}^n \frac{1}{\lambda_1 + \lambda_k}} - \lambda_1 \right)^n \...
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Minimizing $\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}$ for real $a,b,c$ such that $a+b+c=-1$ and $abc\le -3$
Let $a,b,c$ be real number such that $a+b+c=-1$ and $abc\le -3$. Find the minimum value of
$$\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}$$
So, earlier this day I had a competition (it is ...
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