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.

Filter by
Sorted by
Tagged with
2
votes
5answers
50 views

Proving an inequality of three variables

Can someone prove this inequality for the real numbers a,b,c? $$a^2+2b^2+8c^2\geq2a(b+2c)$$ I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $...
2
votes
1answer
57 views

$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$

Let $a,b,c$ be real positive number, $ab+bc+ca=3$. Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$ My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)...
0
votes
0answers
28 views

Logarithm - Minimum Value

I want to find the minimum value of the following expression: $A=\frac{\log_{a}{b}}{a-b+1}+\frac{\log_{b}{c}}{b-c+2}+\frac{\log_{c}{a}}{c-a+3}$ where $a,b,c\in (0,1)$. I can't verify the following ...
1
vote
1answer
41 views

Inequality question involving a cyclic expression and moduli [duplicate]

Consider three real numbers a,b,c such that $a^2+b^2+c^2 =1$. What will be the maximum value of the expression $|a-b| + |b-c| + |c-a|$ ? I tried two approaches. One I used the fact that $a^2+b^2+c^2 =...
2
votes
1answer
69 views

If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$ (Venezuela 1960)

If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $$\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$$ I tried solving it as follows: The condition we want proved is: $\sqrt{3x^2+3-3yz}+\sqrt{3y^...
0
votes
2answers
26 views

Find the mistake (AM GM ineqality)

According to AM GM inequality $$\frac{\left(x^3-\frac{1}{x}-\frac{1}{x}+\frac{2}{x}\right)}{4}\ge\left[(x^3)\left(-\frac{1}{x}\right)\left(-\frac{1}{x}\right)\left(\frac{2}{x}\right)\right]^\frac{1}{4}...
1
vote
2answers
42 views

Proving Young's inequality for n numbers whitout using AM-GM inequality

Let $\alpha_1, ..., \alpha_n \geq 0$, and $p_1,...,p_n \geq 0$ such that $\sum_{i=1}^{n} p_i^{-1} =1$. Show that $\alpha_1...\alpha_n \leq p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n}$. It ...
3
votes
1answer
123 views
+50

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$ The solution goes as follows: $a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}...
2
votes
1answer
59 views

Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Given four positive numbers $a, b, c, d.$ Prove that $$16\max\left \{ a^{3}+ 3bcd, b^{3}+ 3cda, c^{3}+ 3dab, d^{3}+ 3abc \right \}\geq\left ( a+ b+ c+ d \right )^{3}$$ the way I think is using the ...
2
votes
1answer
32 views

Minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ Subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without KKT

I want to minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without using the KKT conditions. The restrain is on regular inequalities like Cauchy or ...
3
votes
1answer
66 views

Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$

given two positive numbers $a, b$ so that $a+ b= 1$ Sui Zhen Lin ; @szl6208 gave a very beautiful proof for the following inequality $$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}+ \sqrt{\frac{4b}{a+ 4}}\leq 1$...
7
votes
0answers
180 views

$\sqrt{\frac{a^2+4bc}{b^2+c^2}}+\sqrt{\frac{b^2+4ac}{a^2+c^2}}+\sqrt{\frac{c^2+4ba}{b^2+a^2}}\ge 2+\sqrt{2}$

Prove that $\forall a,b,c\ge 0$ then $$\sqrt{\frac{a^2+4bc}{b^2+c^2}}+\sqrt{\frac{b^2+4ac}{a^2+c^2}}+\sqrt{\frac{c^2+4ba}{b^2+a^2}}\ge 2+\sqrt{2}$$ I have some ideas but they didn't lead to simpler ...
1
vote
3answers
47 views

Supremum of $x^2y^2(x^2+y^2)$

Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is $3n^2$ ...
4
votes
2answers
73 views

How can we prove the following product-sum inequality involving e?

Take any sequence $x_1, \dots, x_k$ of positive reals such that $\sum_{i=1}^{k-1} x_i < 1$ and $\sum_{i=1}^k x_i \ge 1$. I suspect that the following inequality holds: $$ \prod_{i=1}^k (1 + x_i) \...
1
vote
1answer
48 views

Question regarding a proof of a inequality

Given $a,b,c, \in (0,\infty)$, then the following inequality holds $$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$ What I've tried: First, I noticed that we ...
3
votes
0answers
64 views

$\sum\sqrt{\frac{2a}{b+c}}\le\sqrt[3]{9\sum\frac{a}{b}}$

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}\le\sqrt[3]{9\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$$ It is from ...
2
votes
0answers
54 views

When can we deduce $(mA+nB+pC)>(mX+nY+pZ)$ from $(A+B+C)>(X+Y+Z)$?

I am trying to prove an inequality in the form $(mA+nB+pC)>(mX+nY+pZ)$. I can prove the inequality $(A+B+C)>(X+Y+Z)$ and I wonder if there is any condition, under which we can deduce $(mA+nB+...
0
votes
1answer
63 views

Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$

Question:- Let $a,b$ and $c$ be integers, not all equal to $0$. Show that $$\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$$ This problem was proposed in a canadian journal. The presented ...
4
votes
0answers
152 views

Given three natural numbers $x, y, z.$ Prove that $xy+ yz+ zx\neq 462$ and $xyz+ x+ y+ z\neq 1193$

Given three natural numbers $x, y, z.$ Prove that Problem 1. $$xy+ yz+ zx\neq 462$$ without loss of generality, I accept $x:=\min\left \{ x, y, z \right \}\Rightarrow 3x^{2}\leq 462\Rightarrow x\leq ...
3
votes
2answers
60 views

Prove $\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}-\frac{1}{2}(tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}) \ge \sqrt{3}$

Let $ABC$ be a triangle. Prove that: $$\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}-\frac{1}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}\right) \ge \sqrt{3} $$ My attempt: $$P=\frac{1}{sinA}+\...
4
votes
2answers
163 views

Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$

Let $ABC$ be acute triangle. Prove that $$\sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$$ My attempt: $$\sqrt{\tan A} + \sqrt{\tan B}\geq 2\sqrt[4]{\tan A\cdot\tan B}$$ At here I think I need to ...
1
vote
0answers
64 views

Prove that $2^{n}>1+n\sqrt{2^{\left(n-1\right)}}, \ \ \forall n>2$.

this is a question from the concept of $AM\ge GM\ge HM$ , how do i know which number to select for applying the inequality, please help!
2
votes
4answers
187 views

Using AM-GM to find the minimum [closed]

Find the minimum value of $\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}}$ if $x$ and $y$ are positive real numbers such that $x > y$. This is a question from the 22nd Philippine Mathematical ...
1
vote
1answer
45 views

Notation difficulty in AM-GM inequality proof

The proof goes as follows: If $a_1, \dots, a_n$ are positive numbers then: $$ (a_1 + \dots + a_n)/n \geq \sqrt[n]{a_1 \dots a_n} $$ The left-hand side we denote by $A_n$ and the right-hand side by $...
0
votes
0answers
43 views

Bounds on the AM-GM gap?

I have the expression $\frac{1}{2}(a_1+a_2) - \sqrt{a_1 a_2}$ and I need to upperbound it. Note that this is the "slack in the AM-GM inequality". In my specific case, $0\leq a_1,a_2 \leq 1$. ...
-1
votes
0answers
31 views

minimum of this function quartic function

I am struggling to find the minimum of this function without using calculus: $$f(x)=x^4+x\ \text{and}\ g(x)=2x^4+x$$ Using calculus we get for the first a minimum for $x=-\frac{1}{4}$ with $\min=-\...
4
votes
5answers
97 views

If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+)

Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$ I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I ...
2
votes
2answers
104 views

Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$

$x,y,z \in \mathbb R^+$ different than $0$, Prove that : $$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$ My attempt: First, we should prove that : $$2\left(\frac{xy}{z}+\frac{xz}...
7
votes
1answer
99 views

If $x+y+z+a+b+c=1$, $xyz+abc=1/36$, what is the max value of $abz+bcx+cay$?

Question : If $x+y+z+a+b+c=1,xyz+abc=1/36,$ What is the max value of $abz+bcx+cay$? ($x, y, z, a, b, c ∈ R$ (non-negative)) I tried $abz+bcx+cay≥3abc$ (because of AM-GM) to make it similar to $xyz+...
5
votes
0answers
204 views

Existence of a factorisation proof of the AM-GM inequality

Consider the following formulation of the AM-GM Inequalities: $$\large f_n(a_1, a_2, a_3,\cdots,a_n) := \left(\sum_{k=1}^n a_k^n\right) - n \prod_{k=1}^n a_k \geqslant 0$$ As required by the classical ...
2
votes
2answers
48 views

Question on finding maximum of AM GM inequality

How can i solve for minimum value of function $f(x)=\frac{x^2+x+1}{x^2-x+1} $ using AM GM inequality. I am aware about the AM GM inequality, but am not able to split the terms into the correct form in ...
4
votes
3answers
196 views

maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$

Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$ I couldn't proceed much, however I think I got the minimum and maximum case. For ...
2
votes
0answers
70 views

Prove that $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}$ [duplicate]

If $a,b,c$ are positive reals and $abc=1$, Prove that $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}$$ My try: $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\...
10
votes
2answers
994 views

Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$

Given $x,y,z$ are positive numbers such that $$x^2y+y^2z+z^2x=3$$ Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$ My Attempt: $$P= x^3y+y^3z+z^3x - \left( ...
7
votes
0answers
77 views

Prove $\frac{x}{y^3+2xyz} + \frac{y}{z^3+2xyz} + \frac{z}{x^3+2xyz} \gt \frac{15}{2}$ [duplicate]

Given $x,y,z$ is positive number that $x+y+z=1$. Prove that $$ P = \frac{x}{y^3+2xyz} + \frac{y}{z^3+2xyz} + \frac{z}{x^3+2xyz} \gt \frac{15}{2}$$ My attempt: $$2xyz \leq \frac{2}{27}$$ $$ \Rightarrow ...
0
votes
1answer
50 views

Reciprocal inequality

Given that $a, b, c, d$ are greater than zero, prove that $$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} ≥ \frac{64}{a+b+c+d}$$ The first thing that came to mind was QM-AM-GM-HM, but I haven'...
2
votes
4answers
131 views

Sum of square roots inequality

For all $a, b, c, d > 0$, prove that $$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$ The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to ...
0
votes
1answer
72 views

Solving AM-GM without expanding

Prove that $(p^2+p+1)(q^2+q+1)(r^2+r+1)(s^2+s+1) ≥ 81pqrs$ for all $p, q, r, s ≥ 0$. I am not sure how to solve this problem. I know that this is supposed to be an AM-GM question, so I tried expanding ...
0
votes
2answers
40 views

Minimum of a function by inequality

No answers please, hints only! I am asked to use the fact that $$\cfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\cdots a_n}, n\in \mathbb N \\ \text{with equality if and only if } a_1=a_2=\cdots=a_n$$ To ...
2
votes
1answer
31 views

Proving each subsequent term equal to one another in an equation

Q. The maximum value of $\cos x_1\cdot \cos x_2 \cdot \cos x_3...\cos x_n$ under restrictions $0\le x_1,x_2,x_3,...,x_n \le \frac{\pi}{2}$ and $\cot x_1 \cdot \cot x_2 \cdot \cot x_3...\cot x_n=1$. MY ...
2
votes
1answer
88 views

Inequality $\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq \sqrt{3}$.

Recently I answered this Question which required a proof for the upper and lower bound of a given function. While seemingly a calculus question, I realised that it can be transformed into an ...
0
votes
1answer
59 views

Proof of Hölder's inequality by weighted AM-GM inequality

I am going throw my notes, and for the proof of Hölder's inequality, I wrote the following: Theorem. (Hölder) Let $1 \leq q,r \leq \infty $ such that $ \dfrac{1}{q} + \dfrac{1}{r} = 1 $ (here we ...
6
votes
2answers
174 views

Range of $\frac{\cos\theta_1+\cdots+\cos\theta_{10}}{\sin\theta_1+\cdots+\sin\theta_{10}}$ given $\sin^2\theta_1+\cdots+\sin^2\theta_{10}=1$

Given that for $ \theta_i \in \left[0, \dfrac{\pi}{2}\right]$, where $1 \le i \le 10$ , $\sin^2\theta_1+\sin^2\theta_2+\cdots+\sin^2\theta_{10}=1$, find the minimum and maximum value of $$\dfrac{\cos \...
-1
votes
1answer
56 views

$xy\sqrt[3]\frac{x^3+y^3}{2}\leq \frac{(x+y)^3}{8}$ for $x,y>0$ [closed]

If $x,y>0$ then $$xy\sqrt[3]\frac{x^3+y^3}{2}\leq \frac{(x+y)^3}{8}.$$ Is it possible to prove it by AM-Gm inequality? I have tried but cannot succeed.
3
votes
1answer
56 views

Does AM-GM Follow From the Convexity of Some Function

The AM-GM for $n = 2$ is $$\frac{x+y}{2} \ge (xy)^{1/2}$$ This is very easy to prove with algebra. However, I am wondering if there is a proof using convexity. That is, can we find a convex function $...
7
votes
1answer
148 views

Prove inequality with $a;b;c \in R$

Let $a;b;c \in \mathbb{R} $ such that $a+b+c=0$ Prove that: $P=\dfrac{a-1}{a^2+8}+\dfrac{b-1}{b^2+8}+\dfrac{c-1}{c^2+8} \geq -\dfrac{3}{8}$ I tried to do this: $\dfrac{8a-8}{a^2+8}+2+\dfrac{8b-8}{b^2+...
5
votes
1answer
108 views

Prove this inequality (please using AM-GM or Cauchy-Schwarz if possible)

Let $a ; b ; c > 0$ such that $a+b+c=3$ Prove: $P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{6abc}{ab+bc+ca} \geq 5$ I tried: $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{a^2c+b^2a+c^2b}{abc}=\...
3
votes
3answers
131 views

Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$

It is also given that $abc = 1$. I used AM-GM inequality to reach till $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}...
2
votes
3answers
73 views

If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.

If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. I found this in a Facebook group. I start by doing the math in the LHS: $8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) =...
6
votes
0answers
225 views

Inequality: Prove $\sum_{cyc}\frac{a}{ab+3}\leq\frac{3}{4}$ if $a^2+b^2+c^2=3$

If $a,b,c\geq 0$, $a^2+b^2+c^2=3$, then prove that $$\sum_{cyc}\frac{a}{ab+3}\leq\frac{3}{4}$$ It's a little complex if we just use the pqr method. As far as I'm concerned, the part of denominator is ...

1
2 3 4 5
23