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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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Prove $\sqrt{\frac{a}{b^2+bc+c^2}}+\sqrt{\frac{b}{c^2+ca+a^2}}+\sqrt{\frac{c}{a^2+ab+b^2}}\ge 2\sqrt{2}\sqrt{\frac{(ab+bc+ca)}{(a+b)(b+c)(c+a)}}$

For all $a,b,c\ge 0: ab+bc+ca>0$ then prove $$\color{black}{\sqrt{\frac{a}{b^2+bc+c^2}}+\sqrt{\frac{b}{c^2+ca+a^2}}+\sqrt{\frac{c}{a^2+ab+b^2}}\ge 2\sqrt{2}\sqrt{\frac{(ab+bc+ca)}{(a+b)(b+c)(c+a)}}....
Dragon boy's user avatar
3 votes
1 answer
84 views

Understanding an inequality in the proof

I believe my question is on some very elementary computations (i.e., this is probably answerable even if you don't know what is the subgaussian norm $\| \cdot \|_{\psi_2}$ etc. However, do let me know ...
Partial T's user avatar
  • 561
1 vote
3 answers
143 views

How to prove $x^{x/(1+x)}/(1+x)\geq1/2$

I need to show that $\frac{x^{\frac{x}{1+x}}}{1+x} \geq \frac{1}{2}$ for all $x\geq0$. I am fairly certain this statement is true, but have no idea how to go about proving it. I know that it attains a ...
Vance M's user avatar
  • 112
2 votes
3 answers
126 views

How to prove $\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$

If $a,b,c>0$ prove that $$\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$$ It is stronger than the well-known result $$(a+b+...
Sickness's user avatar
2 votes
5 answers
124 views

Prove $3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$

For any $a,b,c\ge 0$ prove that $$3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$$ Equality holds at $a=b=c=1$ or $a=0$ I tried to use AM as $$6\...
Dragon boy's user avatar
0 votes
2 answers
68 views

Maximization question using inequalities (Cauchy-Schwarz, AM-GM)

Consider this system of equations: \begin{equation*} a^2 + b^2 + c^2 + d^2 = 14 \\ 3a + 2b + c + d = 14 \end{equation*} I want to find the maximum value of $d$ given that $a, b, c, d \in \mathbb{R}$. ...
Christopher Miller's user avatar
0 votes
2 answers
70 views

the maximum value of $xy/(x+2y)$, given $x^2+4y^2+2xy=1, x>0, y>0$

My thought is $\frac{xy}{x+2y} = \frac{1}{2}\frac{x*2y}{x+2y}\le \frac{1}{8}\frac{(x+2y)^2}{x+2y}=\frac{x+2y}{8}$. I don't know how to proceed. Please use AM-GM Inequality only.
JASON CHAN's user avatar
1 vote
2 answers
108 views

For $x,y,z$ positive real numbers $3x+4y+7z=1$. Find minumum integer value of $1/x+1/y+1/z$

For $x,y,z$ positive real numbers $3x+4y+7z=1$. Find minimum integer value of $1/x+1/y+1/z$ My solution like this: Using Cauchy-Schwarz inequality $$(3x+4y+7z)(1/x+1/y+1/z)\ge (\sqrt{3}+2+\sqrt{7})^2$$...
matholympicman's user avatar
4 votes
3 answers
93 views

Prove $\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$

Let $a,b,c>0.$ Prove that $$\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$ WLOG, assuming that $a+b+c=1$ and we'...
Dragon boy's user avatar
-1 votes
2 answers
129 views

Prove $\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} },$ if $a+b+c+abc=4.$

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove that $$\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} .}$$ Source: ...
Sickness's user avatar
2 votes
5 answers
113 views

prove that $\frac{1}{a(b + c)^{3}} + \frac{1}{b(c + a)^{3}} + \frac{1}{c(a + b)^{3}} \ge \frac{27}{8}$ given $a^{2} + b^{2} + c^{2} = 1$

\begin{align} \text{Given } a, b, c > 0 \text{ with } a^{2} + b^{2} + c^{2} = 1, \text{ prove that} \\ \frac{1}{a(b + c)^{3}} + \frac{1}{b(c + a)^{3}} + \frac{1}{c(a + b)^{3}} \ge \frac{27}{8}. \...
Martin.s's user avatar
1 vote
2 answers
151 views

Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

I was given some exercises from Math olympiads, and I am stuck with the one below, which seems soluble, yet I can't come up with something that works. Given that $a,b,c>0$ and $abc=1$, prove that $...
Limsup's user avatar
  • 396
2 votes
1 answer
55 views

$n\geq 2$ and $\sum_{i=1}^{n}x_i^2=1 \Rightarrow \sum_{i=1}^{n}\frac{|x_i|}{\sqrt{1-x_i^2}}\geq 2$

The original inequality is expressed as $$\sum_{i=1}^{n}\frac{a_i}{\sum_{k=1}^{n}a_kb_k-a_ib_i}\geq\frac{4}{\sum_{i=1}^{n}b_i}$$ where $a_i, b_i>0$ for all $i\in\{1,2,\ldots,n\}$. To simplify and ...
LuckyJollyMoments's user avatar
4 votes
2 answers
151 views

Inequality $\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt[4]{8}$ on the unit circle

While playing with a few two variable inequality and AM-GM inequality, I have ran into the following puzzle: Question: Show that if $a, b \in (0,1)$ and $a^2+b^2 = 1$, then: $\dfrac{a}{\sqrt{b}} + \...
Wang YeFei's user avatar
2 votes
2 answers
106 views

Prove an inequality, knowing that $x,y,z$ are strictly positive real numbers

question Let x, y and z be strictly positive real numbers. Prove that the inequality holds: $$\frac{x^5+4x^2+3x+8}{(y+1)^2}+\frac{y^5+4y^2+3y+8}{(z+1)^2}+\frac{z^5+4z^2+3z+8}{(x+1)^2} \ge12$$ my idea ...
IONELA BUCIU's user avatar
  • 1,125
0 votes
1 answer
63 views

4-variable inequality

Prove that for positive $a,b,c,d$ $$\prod_{cyc}(b+c+d-2a) \le abcd$$ By taking the fourth root, both sides are less than $\frac{a+b+c+d}{4}$. A common technique used when this happens is to subtract ...
Stevineon's user avatar
  • 175
0 votes
1 answer
140 views

How to prove $\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\ge 2+\sqrt{\frac{2}{3a+3b+3c-2}}$?

Question. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that $$\color{black}{\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\ge 2+\sqrt{\frac{2}{3a+3b+3c-2}}. }$$ I've tried to use Jichen lemma ...
Anonymous's user avatar
3 votes
2 answers
175 views

How to prove $\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge \frac{\sqrt{abc+4}+4\sqrt{ab+bc+ca+4}}{2}.$

Question. If $a,b,c\ge 0: a+b+c=2,$ prove that $$\color{blue}{\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge \frac{\sqrt{abc+4}+4\sqrt{ab+bc+ca+4}}{2}.}$$Equality holds iff $(a,b,c)=\{(0,1,1);(0,0,2)\}.$ ...
Sickness's user avatar
2 votes
2 answers
50 views

Minimum of $\sum_{cyc}\frac1x$, $\sum_{cyc}x=3a>0$, $x,y,z>0$

Let $f(x,y,z)=\frac1x+\frac1y+\frac1z.$ Impose the constraints $x+y+z=3a$ $x,y,z>0$. What is the minumum value of $f$? Method of Lagrange's multipliers give $-\frac1{x^2}=-\frac1{y^2}=-\frac1{z^2}=\...
Bob Dobbs's user avatar
  • 11.5k
2 votes
2 answers
79 views

Max $P= \frac{2ab+3b^2}{(a+3b)^2}+\frac{2bc+3c^2}{(b+3c)^2}+\frac{2ca+a^2}{(c+3a)^2}$

Let: $a,b,c>0$. Find the maximum value of: $$P= \frac{2ab+3b^2}{(a+3b)^2}+\frac{2bc+3c^2}{(b+3c)^2}+\frac{2ca+a^2}{(c+3a)^2}$$ Here are my try: I tried to use tangent line trick, then I got: $$\...
Lục Trường Phát's user avatar
3 votes
1 answer
66 views

Basic Inequality problem related to AM-GM Inequalities

I am looking to prove $$ \frac{ab}{a⁵+b⁵+ab} + \frac{bc}{b⁵+c⁵+bc} + \frac{ac}{a⁵+c⁵+ac} \le1 $$ Where $ a,b,c $ are positive reals with $ abc = 1 $ My initial thought was to apply AM-GM to get $$ a⁵ +...
Aniket Kumar's user avatar
2 votes
2 answers
198 views

If $a^2+ab+bc+ca<0$, show that $a^2<b^2+c^2$

question Let the real numbers $a,b,c$: a) If $a^2+ab+bc+ca<0$, show that $a^2<b^2+c^2$ b) If $b^2+c^2+ab+bc+ca<0$, show that $a^2>b^2+c^2$. idea The first thing that came into my mind is ...
IONELA BUCIU's user avatar
  • 1,125
2 votes
1 answer
67 views

Let $n \geq 3$ be an integer and $k>1$ be a real number.

Let $n \geq 3$ be an integer and $k>1$ be a real number. Consider the sequence $x_1 \geq x_2 \geq x_3 \geq .... \geq x_n$ be positive real numbers. Prove that $ \dfrac{x_1+kx_2}{x_2+x_3}+\dfrac{x_2+...
The Revolution's user avatar
0 votes
1 answer
152 views

Let $a, b,c$ be positive real numbers such that $a+b+c=1,$ prove $(a + b) ^ 2 (1 + 2c)(2a + 3c)(2b + 3c) \ge54abc$

Let $a, b,c$ be positive real numbers such that $a+b+c=1,$ prove $(a + b) ^ 2 (1 + 2c)(2a + 3c)(2b + 3c) \ge54abc$ I think it's to be solved by AM-GM inequality but I am not getting to the final ...
Sloth28's user avatar
  • 15
0 votes
0 answers
50 views

Proving Inequality Involving Sums and Square Roots with Given Conditions

Question $$\text{Let } b_{i}\wedge a_{i}>0 \text{ where } i\in{1,2,3,\ldots,n} \nonumber , \sum_{i=1}^{n}(b_{i}) = \lambda \text{ then Prove that} \nonumber \frac{\lambda-(b_{1}+b_{2})}{(b_{1}+b_{...
Martin.s's user avatar
2 votes
7 answers
551 views

Minimum value of expression $\displaystyle \sqrt{16b^4+(b-33)^2}$

Finding point $P(a,b)$ on parabola $x=4y^2$ whose distance from the point $Q(0,33)$ is minimum and also find that minimum distance What I try : Let coordinate of point $P$ be $(4b^2,b)$ because point ...
jacky's user avatar
  • 5,331
2 votes
2 answers
111 views

How to prove this inequality $\sum_{r=1}^{n}a_{r}\sqrt{\frac{n-1}{1-a_{r}}}\ge\sum_{r=1}^{n}\sqrt{a_{r}}$

Let $({a_{r}})_{r=1}^{n}$ be a sequence of $n$ positive real numbers that sum to 1. Prove that for all $n>1$ :\begin{align} \sum_{r=1}^{n}a_{r}\sqrt{\frac{n-1}{1-a_{r}}}&\ge\sum_{r=1}^{n}\sqrt{...
Martin.s's user avatar
2 votes
1 answer
49 views

Seeking Clarification on an Inequality Involving Roots and Products

Question If $a_j\gt 0$, $j=1,\ldots,m$ and $x_i\ge 0$ $i=1,\ldots,n.\;$ then prove that \begin{align} \sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}&\ge\frac{1}{n}\sum_{i=1}...
Martin.s's user avatar
0 votes
2 answers
102 views

If $a,b,c > 0$ prove $\frac{(b+c-a)^2}{a}+\frac{(a+c-b)^2}{b}+\frac{(a+b-c)^2}{c}\ge\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}+\frac{a^2+b^2}{a+b}$

$a,b,c > 0$ $$\frac{(b+c-a)^2}{a}+\frac{(a+c-b)^2}{b}+\frac{(a+b-c)^2}{c}\ge\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}+\frac{a^2+b^2}{a+b}$$ I have used Tittu here with the second and the third ...
FabDust's user avatar
  • 195
1 vote
1 answer
112 views

$a,b,c ∈ \mathbb R^+$ and $a+b=1$ then find max value of $a^4b + ab^4$

$a,b,c ∈ \mathbb R^+$ and $a+b=1$ then find max value of $a^4b + ab^4$ So through AM-GM, I found : $$\frac{a+b}{2}\ge\sqrt{ab}\implies ab \le 1$$ Now I have, $$ab(a^3+b^3)\le\frac{1}{4}(a+b)(a^2+b^2-...
FabDust's user avatar
  • 195
-1 votes
2 answers
119 views

If $a,b,c ∈ \mathbb R^+$ prove $(\frac{a}{b+c} + \frac{1}{2})(\frac{b}{c+a} + \frac{1}{2})(\frac{c}{a+b} + \frac{1}{2}) \ge 1$

$a,b,c ∈ \mathbb R^+$ $$(\frac{a}{b+c} + \frac{1}{2})(\frac{b}{c+a} + \frac{1}{2})(\frac{c}{a+b} + \frac{1}{2}) \ge 1$$ For this one I have little to no idea how to solve it. I tried some AM-GM and ...
FabDust's user avatar
  • 195
0 votes
2 answers
123 views

Prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\geq \sqrt{6x+6y+6z}$ [closed]

Prove, that for all positive $x,y,z$ following inequality holds $$\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\geq \sqrt{6x+6y+6z}$$ I have attempted several approaches to prove this inequality, including ...
Darek's user avatar
  • 336
1 vote
1 answer
158 views

Prove $\frac{a}{\sqrt{5-4bc}} + \frac{b}{\sqrt{5-4ac}} + \frac{c}{\sqrt{5-4ab}} \ge 1$ With the condition $a+b+c=3$

$$\frac{a}{\sqrt{5-4bc}} + \frac{b}{\sqrt{5-4ac}} + \frac{c}{\sqrt{5-4ab}} \ge 1$$ With the condition $a+b+c=3$ I am thinking about some type of Cauchy-Schwarz inequality here, but not quite sure how ...
FabDust's user avatar
  • 195
3 votes
2 answers
119 views

What is $(3 n+1)+(3 n+2)+\ldots+(5 n)+(5 n+1)=$ equal to?

I'm trying to prove this inequality$\frac{1}{2}<\frac{1}{3 n+1}+\frac{1}{3 n+2}+\ldots+\frac{1}{5 n}+\frac{1}{5 n+1}<\frac{2}{3}$, and i came across some difficulties with arithmetic progression ...
SysRq308's user avatar
  • 153
0 votes
0 answers
54 views

inequality about elementary symmetric polynomials with real variables

Fix $k\in \mathbb{N}$. Suppose $y_1,\ldots, y_{2k}$ are real numbers. Then $$\left(\sum_{1\le i_1<i_2<\cdots<i_k\le 2k} y_{i_1}y_{i_2}\cdots y_{i_k}\right)^2 \ge \binom{2k}{k}^2y_1y_2\cdots ...
Sayan's user avatar
  • 2,688
-1 votes
1 answer
63 views

Prove $\frac{a+b}{c} + \frac{b+c}{a}+\frac{a+c}{b} + 6 \ge 2\sqrt2(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$ when $a + b + c = 1$

$$\frac{a+b}{c} + \frac{b+c}{a}+\frac{a+c}{b} + 6 \ge 2\sqrt2(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$$ The condition set for this is : $a + b + c = 1$ From that condition I ...
FabDust's user avatar
  • 195
0 votes
0 answers
29 views

Homogeneous Inequality problem unsolved

Let ${x,y,z}$ be positive real numbers. Prove that $\sum\limits_{cyc}\frac{x^2}{yz+\frac{x^4}{y^2}+\frac{x^4}{z^2}} \leq 1$ Here’s my try: By A.M.-G.M. we know that $\frac{yz}{2}+\frac{yz}{2}+\frac{x^...
UWU11's user avatar
  • 137
0 votes
0 answers
48 views

Show that $\sqrt{\frac{a}{b+c}+ \frac{b}{c+a}}+ \sqrt{\frac{b}{c+a}+ \frac{c}{a+b}} + \sqrt{\frac{c}{a+b}+ \frac{a}{b+c}} \ge 3.$

Suppose that $a,b,c>0$, show that $$\sqrt{\frac{a}{b+c}+ \frac{b}{c+a}}+ \sqrt{\frac{b}{c+a}+ \frac{c}{a+b}} + \sqrt{\frac{c}{a+b}+ \frac{a}{b+c}} \ge 3.$$ Appreciate any advice!
Steven Lu's user avatar
  • 1,037
1 vote
1 answer
127 views

Find the min value of $\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}$ when ${x^2} + {y^2} + {z^2} = 1$. [duplicate]

I don't have problems usually with these types of inequalities. They usually have a simple trick where you either take them two by two or else to get the question. Or factor out $x + y + z$ etc. ...
FabDust's user avatar
  • 195
0 votes
0 answers
75 views

Proving an inequality using AM/HM or other inequalities

Let x, y, z be real positive numbers, which are generated from the Ravi transformation of a triangle. Use AM-HM or any other inequality to prove that $$\sqrt{x^2 + y (2 x + 2 z) + 2 y^2 + z^2}<2x+...
Superunknown's user avatar
  • 2,877
0 votes
2 answers
81 views

Logarithmic inequality calculation

Find least integral value of n such that $$\left(\frac{\root \of 5+1}{2}\right)^n > 100$$ I want to solve this without the use of a calculator or logarithmic tables. I tried using AM-GM, such that ...
acelixis's user avatar
  • 309
1 vote
0 answers
69 views

Can we prove Am-GM Inequality using Karamata’s Inequality?

Recently I was trying to prove AM-GM Inequality using several different methods. When it came to Karamata’s Inequality, the proof that flashed through my mind was so simple and amazing, but it came ...
Arachnephob1a's user avatar
1 vote
2 answers
133 views

Seeking maximum parameter value so that inequality always holds true [closed]

What is the largest positive integer $n$ such that $$\frac{a^2}{\frac{b}{29}+\frac{c}{31}} + \frac{b^2}{\frac{c}{29}+\frac{a}{31}} + \frac{c^2}{\frac{a}{29}+\frac{b}{31}} \;\ge\; n\,(a+b+c)$$ holds ...
acelixis's user avatar
  • 309
1 vote
2 answers
111 views

Minimum value of $\frac{\sin x}{4} + \frac{1}{\sin x}$? $x \in (0, \pi/2)$

Question. What is the minimum value of $$ \frac{\sin x}{4} + \frac{1}{\sin x}$$ subject to $x \in (0, \pi/2)$? Applying AM-GM inequality, we get $$ \frac{\sin x}{4} + \frac{1}{\sin x} \geq 1. $$ ...
Tirthik's user avatar
  • 21
3 votes
2 answers
343 views

Muirhead proof $\implies$ AM-GM proof

It is a common sentiment among people involved with maths Olympiads that Muirhead's inequality should only be used to find if an AM-GM proof exists. A common bruteforce technique with inequalities is ...
Eric's user avatar
  • 343
0 votes
2 answers
65 views

How to prove $\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}+2\left[a+b-c(1-\sqrt{3})\right]\ge3\left(\frac{1}{\sqrt[3]{a+c}}+\frac{1}{\sqrt[3]{b+c}}\right).$?

Let $a,b\ge\dfrac{c}{2}\ge0: (c+a)(c+b)>0$. Prove that: $$\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}+2\left[a+b-c(1-\sqrt{3})\right]\ge3\left(\frac{1}{\sqrt[3]{a+c}}+\frac{1}{\sqrt[3]{b+c}}\right).$$ I ...
Anonymous's user avatar
1 vote
4 answers
87 views

Help to prove $\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}$

If $a\ge b>c\ge 0,$ then prove $$\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}.$$ When does equality hold ? I tried to use Cauchy-Schwarz $$(a+b+c)^2\geq 2(a^2+b^2)+13c^2-...
Anonymous's user avatar
2 votes
2 answers
104 views

How to prove $a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\sqrt[3]{abc}\ge 7\sqrt[7]{\frac{abc(a+b)(b+c)(c+a)(a+b+c)}{24}}$?

If $a,b,c\ge 0,$ then prove $$a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\sqrt[3]{abc}\ge 7\sqrt[7]{\frac{abc(a+b)(b+c)(c+a)(a+b+c)}{24}}$$ I found a problem here and try to use AM-GM but it didn't work. By ...
Dragon boy's user avatar
0 votes
1 answer
65 views

If $a,b,c\in R: a+b+c=3,$ find maximum $\min\{a,b,c\}+ \sqrt[3]{abc}-\max\{a,b,c\}..$

If $a,b,c\in R: a+b+c=3,$ find the maximum $$P=\min\{a,b,c\}+ \sqrt[3]{abc}-\max\{a,b,c\}. $$ Some thoughts. I think we just need to prove $\sqrt[3]{abc}\le 1,$ which is true by AM-GM for $a,b,c\ge 0.$...
Anonymous's user avatar
0 votes
2 answers
50 views

Find the maximum $P=\frac{bc}{\sqrt{(1-a)^3(a+1)}}+\frac{ca}{\sqrt{(1-b)^3(b+1)}}+\frac{ab}{\sqrt{(1-c)^3(c+1)}},$ when $a+b+c=1$

Let $a,b,c>0: a+b+c=1.$ Find the maximum$$P=\frac{bc}{\sqrt{(1-a)^3(a+1)}}+\frac{ca}{\sqrt{(1-b)^3(b+1)}}+\frac{ab}{\sqrt{(1-c)^3(c+1)}}.$$ When $a=b=c=\dfrac{1}{3}$ P attained the value $\dfrac{3\...
Anonymous's user avatar

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