cyc sum (a+1)\sqrt{2a(1-a)} \geq 8(ab+bc+ca)

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Amir Hossein

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Amir Hossein

#1 h Jun 25, 2018, 9:31 PM • 5 Y

Y by Devastator, jhu08, Iora, Adventure10, Mango247

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove that
$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$ Also, find the values of $a,b,c$ for which the equality happens.

This post has been edited 1 time. Last edited by Amir Hossein, Jun 25, 2018, 9:35 PM

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DerJan

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DerJan

#2 h Jun 25, 2018, 9:51 PM • 7 Y

Y by Amir Hossein, Devastator, Ankhongu, jhu08, ehuseyinyigit, Adventure10, Mango247

Note that
$\begin{align*} (3a-1)^2 &\geqslant 0 \\ \Leftrightarrow (a+1)^2 &\geqslant 8a(1-a) \\ \Rightarrow (a+1)^2(2a(1-a)) &\geqslant 16a^2(1-a)^2 \\ \Rightarrow (a+1)\sqrt{2a(1-a)} &\geqslant 4a(1-a) = 4a(b+c). \end{align*}$ Thus, we have
$LHS \geqslant 4a(b+c)+4b(c+a)+4c(a+b) = RHS.$

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DerJan

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DerJan

#3 h Jun 25, 2018, 9:56 PM • 3 Y

Y by Amir Hossein, jhu08, Adventure10

Equality at $(a,b,c)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ .

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L3435

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L3435

#4 h Jun 25, 2018, 10:08 PM • 6 Y

Y by Amir Hossein, Devastator, Ankhongu, jhu08, Adventure10, Mango247

$\frac{(2a)+(b+c)}{2}\geq\sqrt{2a(b+c)}$ $\Rightarrow (2a+b+c)\sqrt{2a(b+c)}\geq 4a(b+c)$ Add the cyclic permutations to get the desired inequality. Equality at $2a=b+c$ , $2b=a+c$ and $2c=a+b$ , so $a=b=c$

This post has been edited 1 time. Last edited by L3435, Jun 25, 2018, 10:09 PM

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sqing

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sqing

#5 h Jun 26, 2018, 12:11 AM • 6 Y

Y by Devastator, Amir Hossein, jhu08, Adventure10, Mango247, AylyGayypow009

Amir Hossein wrote:

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove that
$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$ Also, find the values of $a,b,c$ for which the equality happens.

See also here.

Attachments:

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Devastator

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Devastator

#6 h Jun 26, 2018, 12:45 AM • 4 Y

Y by Amir Hossein, jhu08, Adventure10, Mango247

A Jensen's solution cause I am not that good at using AM-GM
Hi!
Is this also gonna be compiled for the contest collections?

This post has been edited 2 times. Last edited by Devastator, Jun 26, 2018, 12:49 AM
Reason: Hi!

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sqing

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sqing

#7 h Jun 26, 2018, 1:31 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Amir Hossein wrote:

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove that
$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$ Also, find the values of $a,b,c$ for which the equality happens.

See also here
https://artofproblemsolving.com/community/c6h1425240p8027908
Let $a,b,c$ be positive real numbers such that $a+b+c=1.$ Prove thhat $\frac{4\sqrt{2}}{3}\ge (1+a)\sqrt{a(1-a)}+(1+b)\sqrt{b(1-b)}+(1+c)\sqrt{c(1-c)} \ge 4\sqrt{2}(ab+bc+ca).$

This post has been edited 1 time. Last edited by sqing, Jun 26, 2018, 2:02 AM

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Amir Hossein

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Amir Hossein

#8 h Jun 26, 2018, 1:49 AM • 4 Y

Y by Devastator, jhu08, Adventure10, Mango247

Devastator wrote:

Is this also gonna be compiled for the contest collections?

Greece JBMO TST 2017.

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277546

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277546

#9 h Jun 26, 2018, 2:12 AM • 3 Y

Y by Amir Hossein, jhu08, Adventure10

My solution:
Notice $2a+(1-a)=a+1 \ge 2\sqrt{2a(1-a)}$ by AM-GM. So the inequality is equivalent to $(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq \sum_{cyc} 2(2a(1-a))=\sum_{cyc} 4a(1-a)= 8\sum_{cyc} ab.$

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TheKingAleks

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TheKingAleks

#10 h May 16, 2019, 8:26 PM • 2 Y

Y by jhu08, Adventure10

sqing wrote:

Amir Hossein wrote:

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove that
$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$ Also, find the values of $a,b,c$ for which the equality happens.

See also here.

Sorry to ask but can you explain to me
Why does the last equality work ?
from $\sqrt{2a(b+c)}\sqrt{2a(b+c)}$ to $4(ab+ac)$ isn't it $2(ab+ac)$

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jasperE3

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jasperE3

#11 h Aug 18, 2021, 5:14 AM • 1 Y

Y by jhu08

The main claim is $(a+1)\sqrt{2a(1-a)}\ge4a(1-a)$ , after which the problem is trivial. Indeed, it's true iff:
$2a(a+1)^2(1-a)\ge16a^2(1-a)^2,$ $\Leftrightarrow(a+1)^2\ge8a(1-a),$ $\Leftrightarrow 9a^-6a+1\ge0,$ $\Leftrightarrow(3a-1)^2\ge0.$
For the upper bound of $\text{LHS}\le\frac83$ , we can use Jensen. The function $f(x)=(x+1)\sqrt{2x(1-x)}$ is concave since $f''(x)=\frac{8x^3-12x^2+3x-1}{2(x(1-x))^{3/2}\sqrt2}$ and $8x^3-12x^2+3x-1\le8x^2-12x^2+3x-1=-\left(x-\frac38\right)^2-\frac7{64}<0$ . Then:
$f(a)+f(b)+f(c)\le3f\left(\frac{a+b+c}3\right)=\frac83.$

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Pseudo_Matter

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Pseudo_Matter

#12 h Apr 18, 2025, 1:54 PM

Y by

Let a + b + c = 1.

We begin with the expression:
$(a + 1)\sqrt{2a(1 - a)}$
Note that:
$(2a + b + c)\sqrt{2a(b + c)} = (a + 1)\sqrt{2a(1 - a)}$
Now consider the inequality:
$\frac{(2a + b + c)}{2} \geq \sqrt{2a(b + c)} \Rightarrow (2a + b + c) \sqrt{2a(b + c)} \geq 4a(b + c)$
Similarly, we have:
$(2b + a + c)\sqrt{2b(a + c)} \geq 4b(a + c)$ $(2c + a + b)\sqrt{2c(a + b)} \geq 4c(a + b)$
Adding all three inequalities:
$(a + 1)\sqrt{2a(1 - a)} + (b + 1)\sqrt{2b(1 - b)} + (c + 1)\sqrt{2c(1 - c)} \geq 8(ab + bc + ca)$
Hence proved.
Equality holds when 2a=b+c
2b=c+a
2c=a+b
Solving this we get a=b=c

This post has been edited 2 times. Last edited by Pseudo_Matter, Apr 18, 2025, 1:56 PM
Reason: Mujhe kya pta

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AylyGayypow009

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AylyGayypow009

#13 h May 11, 2025, 9:43 AM • 2 Y

Y by GayypowwAyly, Foden_phil

sqing wrote:

Amir Hossein wrote:

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove that
$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$ Also, find the values of $a,b,c$ for which the equality happens.