Simple cube root inequality [Taiwan 2014 Quizzes]

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Simple cube root inequality [Taiwan 2014 Quizzes]

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#1 h Jul 18, 2014, 7:33 PM • 8 Y

Y by megarnie, jhu08, HamstPan38825, centslordm, HWenslawski, prMoLeGend42, Adventure10, Sedro

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

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arqady

#2 h Jul 18, 2014, 7:48 PM • 4 Y

Y by jhu08, HWenslawski, Adventure10, Mango247

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

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sqing

#3 h Jul 19, 2014, 8:05 AM • 6 Y

Y by yassinelbk007, kiyoras_2001, jhu08, Saving_Light, Adventure10, Mango247

$3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

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#4 h Jul 19, 2014, 1:07 PM • 3 Y

Y by jhu08, Adventure10, Mango247

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sqing

#5 h Jul 22, 2014, 8:56 AM • 3 Y

Y by jhu08, Adventure10, Mango247

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

Prove that for positive reals $a$ , $b$ we have
$2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}.$
Prove that for positive reals $a$ , $b$ , $c$ , $d$ we have
$4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.$

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#6 h Jul 22, 2014, 9:59 AM • 2 Y

Y by jhu08, Adventure10

sqing wrote:

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

Prove that for positive reals $a$ , $b$ we have
$2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}.$
Prove that for positive reals $a$ , $b$ , $c$ , $d$ we have
$4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.$

hello, your inequality is equivalent to
$\frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0$
after two times squaring.
Sonnhard.

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sqing

#7 h Jul 22, 2014, 10:15 AM • 2 Y

Y by jhu08, Adventure10

Dr Sonnhard Graubner wrote:

sqing wrote:

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

Prove that for positive reals $a$ , $b$ we have
$2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}.$
Prove that for positive reals $a$ , $b$ , $c$ , $d$ we have
$4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.$

hello, your inequality is equivalent to
$\frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0$
after two times squaring.
Sonnhard.

$2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}\iff \frac{1}{4}(a-b)^2(49a^2-2ab+49b^2)\geq 0.$
Thanks.

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sqing

#8 h Jul 24, 2014, 5:50 AM • 3 Y

Y by jhu08, Adventure10, Spiritualsociopath_x

sqing wrote:

$3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

$3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\iff a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc.$

$\frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}=\frac{\sqrt[3]{abc} +\sqrt[3]{abc} +\cdots+\sqrt[3]{abc} +\sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}$

$\le \sqrt[3]{\frac{abc+abc+\cdots+abc+\frac{a^ 3+b^3+c^3}{3}}{9}}= \sqrt[3]{\frac{8abc+\frac{a^ 3+b^3+c^3}{3}}{9}}$

$\Rightarrow 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

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#9 h Jul 24, 2014, 8:00 AM • 3 Y

Y by jhu08, Adventure10, Mango247

sqing wrote:

sqing wrote:

$3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

$3(a+b+c) \ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\iff a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc.$

$\frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}=\frac{\sqrt[3]{abc} +\sqrt[3]{abc} +\cdots+\sqrt[3]{abc} +\sqrt[3]{\frac{a^ 3+b^3+c^3}{3}}}{9}$

$\le \sqrt[3]{\frac{abc+abc+\cdots+abc+\frac{a^ 3+b^3+c^3}{3}}{9}}= \sqrt[3]{\frac{8abc+\frac{a^ 3+b^3+c^3}{3}}{9}}$

$\Rightarrow 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

$9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

By Holder,
${\underbrace{(1+1+\cdots +1)}_{9}}^{2/3}\left(8abc+\frac{a^3+b^3+c^3}{3}\right)^{1/3}\ge 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$
$\Rightarrow 9^{2/3}.9^{1/3}\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

So, it is done

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arqady

#10 h Jul 27, 2014, 5:17 AM • 2 Y

Y by jhu08, Adventure10

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=228831

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bitrak

#11 h Nov 1, 2015, 8:37 AM • 6 Y

Y by Nguyenngoctu, rakhmonfz, raven_, jhu08, Adventure10, ehuseyinyigit

My solution

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bitrak

#12 h Nov 1, 2015, 8:39 AM • 3 Y

Y by jhu08, Adventure10, Mango247

This inequality is <=> with :

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TBazar

#13 h Apr 6, 2016, 7:17 AM • 3 Y

Y by jhu08, Adventure10, Mango247

sqing wrote:

Prove that for positive reals $a$ , $b$ , $c$ , $d$ we have
$4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.$

is it true?

This post has been edited 1 time. Last edited by TBazar, Apr 13, 2016, 10:56 AM

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#14 h Apr 6, 2016, 11:08 PM • 2 Y

Y by jhu08, Adventure10

v_Enhance wrote:

Prove that for positive reals $a$ , $b$ , $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}.$

Stronger inequality:
$3\left( {a + b + c} \right) \ge 8\sqrt[3]{{\frac{{\left( {a + b + c} \right)\left( {ab + bc + ca} \right)}}{9}}} + \sqrt[3]{{\frac{{{a^3} + {b^3} + {c^3}}}{3}}}$ .

This post has been edited 1 time. Last edited by Nguyenngoctu, Apr 6, 2016, 11:13 PM

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r3mark

#15 h Apr 6, 2016, 11:37 PM • 2 Y

Y by jhu08, Adventure10

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#16 h Apr 6, 2016, 11:55 PM • 3 Y

Y by jhu08, Adventure10, Mango247

thanks you!

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TBazar

#17 h Apr 13, 2016, 10:57 AM • 3 Y

Y by jhu08, Adventure10, Mango247

sqing wrote:

Prove that for positive reals $a$ , $b$ , $c$ , $d$ we have
$4(a+b+c+d) \ge 15\sqrt[4]{abcd} + \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.$

Is it true. if it is true then how to prove it?

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arqady

#18 h Apr 14, 2016, 1:11 PM • 4 Y

Y by luofangxiang, jhu08, Adventure10, Mango247

By $uvw$ we can show that the following inequality is also true.
Let $a$ , $b$ , $c$ and $d$ be non-negative numbers. Prove that:
$2(a+b+c+d)\geq7\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}+\sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$

This post has been edited 1 time. Last edited by arqady, Apr 14, 2016, 1:16 PM

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booth

#19 h May 27, 2020, 1:39 PM • 1 Y

Y by jhu08

Let $4u=a+b+c+d$ , $6v^2=ab+cd+ac+bd+ad+bc$ , $4w^3=abc+bcd+cda+dab$ and $t^4=abcd$ , then we need to prove
$8u\geq 7w+\sqrt[4]{64u^4-96u^2v^2+18v^4+16uw^3-t^4}$ or
$8u-7w\geq \sqrt[4]{64u^4-96u^2v^2+18v^4+16uw^3-t^4}$ or
$(8u-7w)^4\geq 64u^4-96u^2v^2+18v^4+16uw^3-t^4$ or
$4032u^4+96u^2v^2-18v^4-14336u^3w+18816u^2w^2-10992uw^3+2401w^4+t^4$ .
Since $t^4\geq4uw^3-3v^4$ . thus
$4032u^4+96u^2v^2-18v^4-14336u^3w+18816u^2w^2-10992uw^3+2401w^4+t^4\geq 4032u^4+96u^2v^2-21v^4-14336u^3w+18816u^2w^2-10988uw^3+2401w^4$ .
By Rolle's theorem $(X-a)(X-b)(X-c)(X-d)'=4(X^3-3uX^2+3v^2X-w^3)$ has three real roots. Let $3u=x+y+z$ , $3v^2=xy+yz+zx$ and $w^3=xyz$ , then we need to prove $f(v^2)\geq 0$ , where $f(v^2)$ is concave, which means enough to check when two of variables are equal. WLOG $y=z=1$ . then it's equivalent to
$(x - 1)^2 (1344 x^{10} + 2688 x^9 - 10304 x^8 - 12352 x^7 + 42048 x^6 + 10432 x^5 - 87208 x^4 + 40944 x^3 + 61891 x^2 - 71038 x + 21825)\geq 0$ , which is true.
P.S- in the last part I substituted $x\rightarrow x^3$ .

This post has been edited 2 times. Last edited by booth, May 27, 2020, 3:57 PM

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#20 h Oct 17, 2020, 7:21 PM • 1 Y

Y by jhu08

WLOG let $abc=1$ since this equation is homogeneous. We note that weighted power mean when we set weights $w_1=\frac19$ and $w_2=\frac89$ gets that letting the positive real numbers $a_1=\frac13(a^3+b^3+c^3)$ and $a_2=1$ gets when we have the power-mean number as $1$ vs $\frac13$ , we have the inequality $\frac19 \cdot \frac13(a^3+b^3+c^3) + \frac89 \ge \left[\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89\right]^3$ which means since $\frac19 \cdot \frac13(a^3+b^3+c^3) = \frac{a^3+b^3+c^3+24abc}{27}\le \frac{\sum_{\text{cyc}}{a^3}+3\sum_{\text{sym}}{a^2b}+6abc}{27} = \frac{1}{27}(a+b+c)^3$ by Muirhead's Inequality over $3\sum_{\text{sym}}{a^2b} \ge 3 \cdot \sum_{\text{sym}}{abc} = 18abc$ . As a result, $\frac{(a+b+c)^3}{27} \ge \left[\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89\right]^3$ so we can take the cube root of both sides getting $\frac13(a+b+c) \ge \frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac89=\frac19 \cdot\sqrt[3]{\frac13(a^3+b^3+c^3)} + \frac{8abc}9$ and multiplying both sides by $9$ gets $3(a+b+c) \ge \sqrt[3]{\frac13(a^3+b^3+c^3)} + 8abc$ which is our desired inequality.

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920 posts

Grizzy

#21 h Dec 31, 2020, 4:18 AM • 2 Y

Y by teomihai, jhu08

By Power Mean, we see that

$\begin{align*} \frac{8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}}{9} & \le \sqrt[3]{\frac{8abc + \frac{a^3 + b^3 + c^3}{3}}{9}} \end{align*}$
and so

$\begin{align*} 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} & \le 9 \cdot \sqrt[3]{\frac{8abc + \frac{a^3 + b^3 + c^3}{3}}{9}}\\ & = 3\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\\ & \le 3\sqrt[3]{(a+b+c)^3}\\ & = 3(a+b+c) \end{align*}$
as desired. $\square$

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#22 h Mar 30, 2021, 4:44 PM • 2 Y

Y by blueballoon, jhu08

By weighted power mean with weights $\tfrac19, \tfrac89$ and reals
$\frac{a^3+b^3+c^3}{3}, abc$ we have
$\begin{align*} \frac{\frac{a^3+b^3+c^3}{3}+8abc}{9}&\ge \left(\frac{\sqrt[3]{\frac{a^3+b^3+c^3}{3}} + 8\sqrt[3]{abc}}{9}\right)^3.\\ \end{align*}$ After doing some manipulations, we see it suffices to show
$a^3+b^3+c^3+24abc \le (a+b+c)^3$ or
$18abc \le 3a^2b+3a^2c+3ab^2+3ac^2+3bc^2+3b^2c.$ Since $(2,1,0)\succ(1,1,1)$ by Muirhead, we have
$\sum_{sym}a^2b \ge \sum_{sym}abc=6abc,$ so multiplying be $3$ , we are done. $\blacksquare$

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#23 h Aug 19, 2021, 2:43 PM • 3 Y

Y by hakN, jhu08, Executioner230607

Set $f(x)=\sqrt[3]{x}$ which is clearly concave. By Jensen's, $9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$ so it suffices so show the stronger $\begin{align*}3(a+b+c)&\ge 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}}\\ \iff \frac{(a+b+c)^3}{27}&\ge \frac{\frac{3\cdot 8abc}{3}+\frac{a^3+b^3+c^3}{3}}{9}\\ \iff (a+b+c)^3&\ge a^3+b^3+c^3+24abc\\ \iff a^3+b^3+c^3+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)+6abc &\ge a^3+b^3+c^3+24abc\\ \iff a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\ge 6abc \end{align*}$ which is obvious by Muirhead/AM-GM.

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1757 posts

Taco12

#24 h Sep 15, 2021, 8:44 PM • 1 Y

Y by jhu08

Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$ , and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields

$\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$ .

Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$ , or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$ . This follows directly from Muirhead, so we are done.

This post has been edited 2 times. Last edited by Taco12, Sep 15, 2021, 8:45 PM

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#25 h Sep 16, 2021, 5:11 AM • 1 Y

Y by jhu08

sqing wrote:

Prove that for positive reals $a$ , $b$ we have
$2(a+b)\ge 3\sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}}.$

Good.

Solution

Second is much harder with this method.

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#26 h Oct 6, 2021, 9:36 AM

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Storage.

This post has been edited 1 time. Last edited by geometry6, Jun 3, 2023, 9:05 PM

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#27 h Oct 17, 2021, 6:36 PM

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Setting $s=a+b+c, q=\sqrt{3ab+3bc+3ca}, p=3\sqrt[3]{abc}$ (the idea is to use the three symmetric expressions for $a,b,c$ , so that $s=p=q$ if $a=b=c$ and then get rid of the cube root, using $s,q,p$ ). We can transfrom the inequality as follows:
$3(a+b+c)=3s,\quad 8\sqrt[3]{abc}=\frac{8}{3}p,\quad a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ca)(a+b+c)+3abc=s^3-q^2s+\frac{1}{9}p^3$ $\implies 3(a+b+c)\geq 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\iff 3s-\frac{8}{3}p\geq \sqrt[3]{\frac{s^3-q^2s+\frac{1}{9}p^3}{3}}$ $\iff 9s-8p\geq \sqrt[3]{9s^3-9q^2s+p^3}\iff 729s^3-1944s^2p+1728sp^2-512p^3\geq 9s^3-9q^2s+p^3$ $\iff 720s^3+9q^2s+1728sp^2-1944s^2p-513p^3\geq 0$ However, by AM-GM we have that $s\geq q\geq p$ and we can finish the solution substituting $q$ with $p$ and factoring the polynomial as we know that $s=p$ will be a root as obviously the inequality case is achieved if $a=b=c\implies s=p$ :
$720s^3+9q^2s+1728sp^2-1944s^2p-513p^3\geq 720s^3+9p^2s+1728sp^2-1944s^2p-513p^3=9(s-p)(4s-3p)(20s-19p)\geq 0$ The last inequality follows from $s\geq p>0\implies 4s-3p>0, 20s-19p>0, s-p\geq 0$ , thus we've solved the problem! (also we know that the inequality reaches equality iff $s=p\implies a=b=c$ )

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gambi

#28 h Oct 30, 2021, 10:30 PM

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Consider the following set of $9$ elements:
$\left\{8\sqrt[3]{abc},8\sqrt[3]{abc},\dots,8\sqrt[3]{abc}, \thickspace 8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right\} \, ,$ Applying the inequality between the cubic and arithmetic means to the elements of the set, one gets
$\frac{64\sqrt[3]{abc}+8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}}{9}\leq \sqrt[3]{\frac{8\cdot (8\sqrt[3]{abc})^3+\left(8\cdot \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right)^3}{9}},$ which is equivalent to
$8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\leq 3\cdot \sqrt[3]{24 \, abc+a^3+b^3+c^3}$
Hence in order to prove the heading it suffices to show
$3(a+b+c)\geq 3\cdot \sqrt[3]{24 \, abc+a^3+b^3+c^3}$ And this is clear, because
$(a+b+c)^3\geq a^3+b^3+c^3+24abc \iff 3\cdot (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)\geq 18abc \iff$ $\iff \frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\geq 6,$ which is true by AM-GM.

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#29 h Oct 30, 2021, 10:46 PM

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Can we generalize this inequality into $n$ -variables?

Let $n\geq 2$ be an integer and $a_1,a_2,\cdots,a_n$ be positive reals. Find min $\lambda$ such that

$\sum_{i=1}^n a_i \geq \lambda \sqrt[n]{\prod_{i=1}^n a_i} + (1 - \lambda) \sqrt[n]{\frac{1}{n}\sum_{i=1}^n a_i^n}$

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#30 h Nov 14, 2021, 2:41 PM

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Notice that since $f(x)=\sqrt[3]{x}$ is concave so by Jensen's we have $\frac{8f(abc)+f\left(\frac{a^3+b^3+c^3}{3}\right)}{9}\le f\left(\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}\right).$ Then we have
$\begin{align*} 8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\le 9\sqrt[3]{\frac{8abc+\frac{a^3+b^3+c^3}{3}}{9}} \\ &= 3\sqrt[3]{a^3+b^3+c^3+24abc} \\ &\le 3\sqrt[3]{(a+b+c)^3} \\ &= 3(a+b+c). \end{align*}$

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#31 h Dec 29, 2021, 10:10 AM

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$(a+b+c)^3\geq a^3+b^3+c^3+24abc$ $a^3+b^3+c^3=3x^3$ $abc=y^3$ $81(x^3+8y^3)\geq (x+8y)^3$ Remaining easy.

This post has been edited 1 time. Last edited by lazizbek42, Dec 29, 2021, 10:14 AM

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#32 h Nov 21, 2022, 10:20 PM

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Taco12 wrote:

Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$ , and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields

$\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$ .

Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$ , or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$ . This follows directly from Muirhead, so we are done.

interesting, but I don't understand why the last inequality ( $a^3+b^3+c^3+24abc \le (a+b+c)^3$ )is what is missing to solve the problem.
anyone could explain me please?

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#33 h Mar 15, 2023, 11:13 PM

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By Power Mean with weights $\frac 89$ and $\frac 19$ , $\left(\frac 89\sqrt[3]{abc} + \frac 19\sqrt[3]{\frac{a^3+b^3+c^3}3}\right)^3 \leq \frac 89 abc + \frac 19 \cdot \frac{a^3+b^3+c^3}3.$ Thus, it suffices to show that $\left(\frac{a+b+c}3\right)^3 \geq \frac 19\left(8abc+\frac{a^3+b^3+c^3}3\right) \iff (a+b+c)^3 \geq 24abc+a^3+b^3+c^3.$ This is evident.

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#34 h May 30, 2023, 12:21 AM

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Note how most of these solutions were just like Evan’s in OTIS Excerpts

Assim584 wrote:

Taco12 wrote:

Weighted power mean with weights $\frac{8}{9}$ and $\frac{1}{9}$ , and 2 numbers $\frac{a^3+b^3+c^3}{3},abc$ yields

$\frac{1}{9}\left(\frac{a^3+b^3+c^3}{3}\right)+\frac{8}{9}(abc) \geq \frac{1}{9}\sqrt{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt{abc}$ .

Thus, we wish to prove $a^3+b^3+c^3+24abc \le (a+b+c)^3$ , or $a^2b+a^2c+ab^2+ac^2+bc^2+b^2c \ge 6abc$ . This follows directly from Muirhead, so we are done.

interesting, but I don't understand why the last inequality ( $a^3+b^3+c^3+24abc \le (a+b+c)^3$ )is what is missing to solve the problem.
anyone could explain me please?

This is a very well known result, and easy to prove. Expanding, it suffices to prove that $3(a^2b+ab^2+b^2c+c^2b+a^2c+c^2a)\geq 3(6abc)$ , which follows immediately from dividing by 3 and AM-GM over all of the terms. Muirhead with (2,1,0) majorizing (1,1,1) would also suffice.

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#35 h Sep 28, 2023, 3:27 AM

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This was rough.

Power mean inequality gives,
$\left( \frac{8}{9} \cdot \sqrt[3]{abc} + \frac{1}{9}\cdot \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \right)^3 \leq \frac{8}{9} \cdot abc + \frac{1}{9} \cdot \frac{a^3+b^3+c^3}{3}.$ Then we wish to show,
$\begin{align*} (a+b+c)^3 &\geq 24abc + a^3+b^3+c^3\\ 3\sum_{sym} a^2b &\geq 18abc \end{align*}$ which is just Muirheads.

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#36 h Oct 1, 2023, 3:19 AM

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By the Power Mean, we have that
$\frac{\frac{a^3+b^3+c^3}{3}+8abc}{9}\geq\left(\frac{\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+8\sqrt[3]{abc}}{9}\right)^3.$ Which is equivalent to
$81(\frac{a^3+b^3+c^3}{3}+8abc)\geq (\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+8\sqrt[3]{abc})^3,$ meaning that we just have to prove that
$27(a+b+c)^3\geq 81(\frac{a^3+b^3+c^3}{3}+8abc),$ or
$(a+b+c)^3 \geq a^3+b^3+c^3+24abc.$ Expanding, this is equivalent to proving that
$a^2b+b^2a+a^2c+c^2a+b^2c+c^2b\geq 6abc,$ which is true by Muirhead's, since $(2,1,0)$ majorizes $(1,1,1)$ , finishing the problem.

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chakrabortyahan

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#37 h Oct 5, 2023, 3:01 PM

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Nice problem.
first we take $f(x) = x^{\frac{1}{3}} , x>0$ Note that $f'= 1/3 x^{-2/3} >0$ and $f" = (1/3)\cdot(-2/3) x^{-5/3} < 0$ and so $f$ is concave on $(0,\infty)$
Dividing both sides by $9$ the inequality is reduced to $\frac{a+b+c}{3} \geq \frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt[3]{abc}$ Now note that by Jensen's inequality, $\frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{8}{9}\sqrt[3]{abc}$ $= \frac{8}{9} f (abc)+\frac{1}{9} f (\frac{a^3+b^3+c^3}{3})$
$\leq f(\frac{a^3+b^3+c^3+24abc}{27})$ Now note that $a^3+b^3+c^3+24abc \leq (a+b+c)^3$ (do AM GM , Muirhead whatever) and $f$ is increasing so $f(\frac{a^3+b^3+c^3+24abc}{27}) \leq f(\frac{(a+b+c)^3}{27}) = \frac{a+b+c}{3}$ $\blacksquare$

This post has been edited 1 time. Last edited by chakrabortyahan, Oct 5, 2023, 3:01 PM

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#38 h Oct 23, 2023, 10:02 PM • 1 Y

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#39 h Jan 24, 2024, 6:36 PM

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Powerful problem,
By Weighted Power Mean with weights $\tfrac{1}{9}$ , and $\tfrac{8}{9}$
$\frac{8abc}{9}+\frac{a^{3}+b^{3}+c^{3}}{27}\geq \left(\frac{8}{9} \sqrt[3]{abc}+ \frac{1}{9} \sqrt[3]{\tfrac{a^3+b^3+c^3}{3}} \right)^{3}$ $3 \sqrt[3]{a^{3}+b^{3}+c^{3}+24abc} \geq 8\sqrt[3]{abc} + \sqrt[3]{\tfrac{a^3+b^3+c^3}{3}}$ So, it suffices to show:
$3(a+b+c) \geq 3 \sqrt[3]{a^{3}+b^{3}+c^{3}+24abc}$ $(a+b+c)^{3} \geq a^{3}+b^{3}+c^{3}+24abc$ Which is obvious by expansion and muirhead $\square$

This post has been edited 1 time. Last edited by bjump, Jan 24, 2024, 6:36 PM

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#40 h Feb 22, 2024, 2:23 PM

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By wieghted power mean we get $\left(\frac{8\sqrt[3]{abc}}{9} + \frac{1}{9} \cdot \frac{\sqrt[3]{a^3+b^3+c^3}}{3}\right)^3 \leq \frac{a^3+b^3+c^3}{27}+\frac{8abc}{9}$ Thus it remains to show that $\frac{a^3+b^3+c^3}{27}+\frac{8abc}{9} \leq \frac{(a+b+c)^3}{27} \Rightarrow a^3+b^3+c^3 +24 abc \leq (a+b+c)^3$ This is obvious from Muirhead.

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#41 h Sep 28, 2024, 4:11 AM

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Apply weighted power mean on $P(1) \geq P(\frac13)$ with weights $\frac89$ and $\frac19$ and positive reals $abc$ and $\frac{a^3+b^3+c^3}{3}$ . After dividing both sides by $9$ , the RHS is then $\sqrt[3]{P(\frac13)}$ so it suffices to prove $\frac{a+b+c}{3} \geq \sqrt[3]{P(\frac13)}$ or $(\frac{a+b+c}{3})^3 \geq P(1)$ by our weighted power mean. Now $P(1) = \frac{24abc+a^3+b^3+c^3}{27}$ so we just need to prove that $(a+b+c)^3 \geq 24abc + a^3+b^3+c^3$ $3 \sum_{\text{sym}} ab^2 \geq 18abc$ after expansion which is true by a simple AMGM.

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#42 h Mar 19, 2025, 3:22 AM

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Power mean

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