Not easy one

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red3

905 posts

red3

#1 h Aug 6, 2011, 2:08 AM • 2 Y

Y by Ji Chen, Adventure10

The positive reals $a,b,c,d$ satisfy $abcd=1$ . Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$ .

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fjwxcsl

1630 posts

fjwxcsl

#2 h Aug 6, 2011, 2:22 AM • 6 Y

Y by rightways, David-Vieta, Adventure10, Mango247, and 2 other users

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{k}{a+b+c+d}\ge \frac{16+k}{4}$
holds for all $a,b,c,d>0,abcd=1$ ,then $k_{max}=k_0=12.4185242...,$ where

$-1938817024-334692352k_0-172032k_0^2+2304512k_0^3+62768k_0^4+729k_0^5=0.$

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DC93

214 posts

DC93

#3 h Aug 6, 2011, 6:17 AM • 4 Y

Y by yilun, Adventure10, Mango247, and 1 other user

It's CGMO2011,p3
I have proved this inequality in a way which is not so nice……

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DC93

214 posts

DC93

#4 h Aug 6, 2011, 6:21 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Try also this one:(my solution work on this too)
$abcd = 1,a,b,c,d > 0$
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{{16}}{{a + b + c + d}} \ge \frac{{40\sqrt 3 }}{9}$
When does the equality hold?

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oldbeginner

3428 posts

oldbeginner

#5 h Aug 6, 2011, 6:25 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Show us your slution, please!

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arqady

30213 posts

arqady

#6 h Aug 6, 2011, 8:28 AM • 16 Y

Y by oldbeginner, Ji Chen, Sutuxam, colorfuldreams, shinichiman, hctb00, hoangpham, champion999, luofangxiang, KastavIvan, Adventure10, Mango247, MS_asdfgzxcvb, and 3 other users

I'll prove a stronger inequality:
Let $a$ , $b$ , $c$ and $d$ are positives such that $abcd=1$ . Prove that:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7$
Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$ .
Hence, $f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$
$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(a+b+c+d)\left(a+3\sqrt[3]{bcd}\right)}\geq$
$\geq\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(\frac{b+c+d}{3}+b+c+d)\left(\frac{b+c+d}{3}+3\sqrt[3]{bcd}\right)}=$
$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}$ .
We'll prove that $\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0$ .
Let $b+c+d=3u$ , $bc+bd+cd=3v^2$ and $bcd=w^3$ .
Hence, $\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0\Leftrightarrow g(v^2)\geq0$ ,
where $g$ is linear increasing function. Hence, $g$ gets a minimal value, when $v^2$ gets a minimal value,
which happens, when two numbers from $\{b,c,d\}$ are equal.
Since $g(v^2)\geq0$ is homogeneous, for the proof of $g(v^2)\geq0$ we need to check only one case:
$c=d=1$ , which after substitution $b=x^3$ gives
$(x-1)^2(2x^7+x^6+18x^5-10x^4-50x^3+36x^2+26x+4)\geq0$ , which is true.
Id est, $f(a,b,c,d)\geq f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)$ and it remains to prove that
$f(a,b,b,b)\geq0$ , where $a=\frac{1}{b^3}$ , which gives
$(b-1)^2(3b^6+6b^5+9b^4-9b^3-5b^2-b+3)\geq0$ , which is true.
By the same way we can check the fjwxcsl's inequality.

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DC93

214 posts

DC93

#7 h Aug 6, 2011, 2:18 PM • 4 Y

Y by Kunihiko_Chikaya, Adventure10, Mango247, and 1 other user

I really want to see a solution without MV method...My solution is not compeletely MV but a bit like that

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mudok

3377 posts

mudok

#8 h Aug 6, 2011, 5:44 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

What is CGMO?
Is official solution also by MV method?

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DC93

214 posts

DC93

#9 h Aug 7, 2011, 1:01 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

mudok wrote:

What is CGMO?
Is official solution also by MV method?

It is China Girls Math Olympiad
I havent seen the official solution yet……

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fjwxcsl

1630 posts

fjwxcsl

#10 h Aug 7, 2011, 4:19 AM • 3 Y

Y by Adventure10 and 2 other users

red3 wrote:

The positive reals $a,b,c,d$ satisfy $abcd=1$ . Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$ .
See here:
http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=96&id=23073

fjwxcsl wrote:

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{k}{a+b+c+d}\ge \frac{16+k}{4}$
holds for all $a,b,c,d>0,abcd=1$ ,then $k_{max}=k_0=12.4185242...,$ where

$-1938817024-334692352k_0-172032k_0^2+2304512k_0^3+62768k_0^4+729k_0^5=0.$ .

fjwxcsl wrote:

Determine the maximum $k_1$ such that
$(\forall{x_1,x_2,x_3,x_4}>0)x_{1}+x_{2}+x_{3}+x_{4}+\frac{4k_1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}} \ge{(4+k_1){\sqrt[4]{x_{1}x_{2}x_{3}x_{4}}}}.$

FROM《fjwxcsl:210 nice symmetric inequalities(4,5), $T_{4.9(1)}$ 》,SEE:

http://www.irgoc.org/viewtopic.php?f=10&t=926&sid=59fe05e18960535ca0c6b19a0394321c&start=0

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red3

905 posts

red3

#11 h Aug 7, 2011, 8:33 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

but what if k>16?

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DC93

214 posts

DC93

#12 h Aug 7, 2011, 8:39 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

red3 wrote:

but what if k>16?

The answer is very ugly

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fjwxcsl

1630 posts

fjwxcsl

#13 h Aug 8, 2011, 12:58 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The positive reals $a,b,c,d,e$ satisfy $abcde=1$ . Prove that

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{16}{{a + b + c + d+e}} \ge{\frac{{41}}{5}}.$

Remark

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Tourish

663 posts

Tourish

#14 h Aug 9, 2011, 2:17 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Here is a similar result, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=354052

Tourish wrote:

$x_1,x_2,\cdots,x_n>0$ ,prove that
$x_1+x_2+\cdots+x_n+\frac{n(n-1)}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}\geq k_n\sqrt[n]{x_1x_2\cdots x_n}$
where
$k_n= \begin{cases} 2n-1,&2\leq n\leq 4\\ \sqrt[n]{\frac{2(n-1)}{n-2-\sqrt{n(n-4)}}}\cdot\frac{2n\left[n^2-2n-(n-1)\sqrt{n(n-4)}\right]}{(n-1)\left[n-\sqrt{n(n-4)}\right]},&n\geq 5 \end{cases}$
where $k_n$ has the best result .

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littletush

761 posts

littletush

#15 h Aug 13, 2011, 5:21 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

just by $EV$ ,one can get it.

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Ghoshadi

925 posts

Ghoshadi

#42 h Jan 25, 2017, 8:27 PM • 1 Y

Y by Adventure10

Kunihiko_Chikaya wrote:

red3 wrote:

The positive reals $a,b,c,d$ satisfy $abcd=1$ . Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$ .

I have just found the geometric proof as follows.

Regrettably, the proof was not perfect, but I dare to remain as memorial.

Kunihiko Chikaya

Take points $A\left(a,\ \frac{1}{a}\right),\ B\left(b,\ \frac{1}{b}\right),\ C\left(c,\ \frac{1}{c}\right),\ D\left(d,\ \frac{1}{d}\right)$ on the hyperbola $y=\frac{1}{x}\ (x>0),$ W.L.O.G. $0<a\leq b\leq c\leq d$ .

Let $K,\ L$ be the midpoints of the line segments $AB,\ CD$ respectively, we have $K\left(\frac{a+b}{2},\ \frac{\frac{1}{a}+\frac{1}{b}}{2}\right),\ L\left(\frac{c+d}{2},\ \frac{\frac{1}{c}+\frac{1}{d}}{2}\right)$ , we have the midpoint $M\left(\frac{a+b+c+d}{4},\ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4}\right)$ of the line segment $KL$ . Let $N\left(\frac{a+b+c+d}{4},\ \frac{1}{\frac{a+b+c+d}{4}}\right).$ Since $y=\frac{1}{x}\ (x>0)$ is a convex and decreasing function, take a point $P$ such that $MP:PN=9:16$ on the line segment $MN$ , denoting by $y_{P}$ the $y$ -coordinate of the point $P$ , we have $y_{N}\leq y_{P}\leq y_{M}\ \cdots (*)$ , then by the AM-GM inequality, from the condition $abcd=1$ , we have $y_N=\frac{1}{\frac{a+b+c+d}{4}}\leq \frac{1}{\sqrt[4]{abcd}}$ , yielding $y_N\leq 1.$ Therefore from $(*)$ , we have $y_P=\frac{16\cdot \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4}+9\cdot \frac{1}{\frac{a+b+c+d}{4}}}{9+16}\geq 1$

$\Longleftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\geq \frac{25}{4}.$

Equality holds when $a=b=c=d=1$ , or all of points $A,\ B,\ C,\ D$ coincide.

$y_p\leq 1$ and $y_p\leq y_n$ does not imply $y_n\geq 1$ , sir.

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mudok

3377 posts

mudok

#43 h Jan 26, 2017, 1:16 PM • 2 Y

Y by Adventure10, Mango247

red3 wrote:

The positive reals $a,b,c,d$ satisfy $abcd=1$ . Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$ .

I tried to find an elementary proof. Used only Schur, AM-GM and factorizations.

After substitutions $a\to \frac{1}{a}, ...$ we need to prove $a+b+c+d+\frac{9}{abc+bcd+cda+dab}\ge \frac{25}{4}$

A weak lemma:

$(a+b+c+d)^3+\frac{128abcd}{a+b+c+d}\ge 24(abc+bcd+cda+dab)$

Proof of the lemma:

WLOG $a+b+c+d=4, \ \ d=min(a,b,c,d)\le 1$ .

We need to prove $32+16abcd\ge 12abc+3d\cdot 4(ab+bc+ca)$

$\Leftarrow 32+16abcd\ge 12abc+3d ((a+b+c)^2+\frac{9abc}{a+b+c}) \ \ \ \ \ \ \text{(by Schur's inequality)}$

$\Leftarrow 32\ge 12abc(1-d)+3d(a+b+c)^2+ abc( \frac{27}{a+b+c}-4)d \ \ \ \ \ \text{By AM-GM:}$

$\Leftarrow 32\ge \frac{4(a+b+c)^3}{9} (1-d)+3d(a+b+c)^2+ \frac{(a+b+c)^3}{27} \cdot( \frac{27}{a+b+c}-4)d$

$\Leftarrow 32\ge \frac{4(4-d)^3}{9} (1-d)+3d(4-d)^2+ \frac{(4-d)^3}{27} \cdot( \frac{27}{4-d}-4)d$

$(d-1)^2(6+4d-d^2)\ge 0$ . Proof of the lemma is finished.

Let $a+b+c+d=x\ge 4$ . By the lemma we have $\frac{x^4+128}{24x}\ge abc+bcd+cda+dab$ .

So it is enough to prove

$x+\frac{9\cdot 24x}{x^4+128}\ge \frac{25}{4} \ \ \ \Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 800)\ge 0$

$\Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 704)\ge 0$

$\Leftarrow \ \ (x - 4)^2 (4 x^3 + 7 x^2 - 8 x - 176)\ge 0$

Which is obvious due to $x\ge 4$ .

This post has been edited 3 times. Last edited by mudok, Jan 26, 2017, 1:47 PM

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sqing

41825 posts

sqing

#44 h Jun 3, 2018, 12:39 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Let $a$ , $b$ , $c$ and $d$ are positives such that $abcd=1$ . Prove that:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7$

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$

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sqing

41825 posts

sqing

#45 h Oct 28, 2018, 2:10 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{n^2(\sqrt[n]{2n}-1)}{a_1+a_2+\cdots+a_n}\geqslant n\sqrt[n]{2n}.$ (Jichen)
p/4089769260

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SRIDARA

642 posts

SRIDARA

#46 h Oct 28, 2018, 2:14 AM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

red3 wrote:

The positive reals $a,b,c,d$ satisfy $abcd=1$ . Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$ .

I tried to find an elementary proof. Used only Schur, AM-GM and factorizations.

After substitutions $a\to \frac{1}{a}, ...$ we need to prove $a+b+c+d+\frac{9}{abc+bcd+cda+dab}\ge \frac{25}{4}$

A weak lemma:

$(a+b+c+d)^3+\frac{128abcd}{a+b+c+d}\ge 24(abc+bcd+cda+dab)$

Proof of the lemma:

WLOG $a+b+c+d=4, \ \ d=min(a,b,c,d)\le 1$ .

We need to prove $32+16abcd\ge 12abc+3d\cdot 4(ab+bc+ca)$

$\Leftarrow 32+16abcd\ge 12abc+3d ((a+b+c)^2+\frac{9abc}{a+b+c}) \ \ \ \ \ \ \text{(by Schur's inequality)}$

$\Leftarrow 32\ge 12abc(1-d)+3d(a+b+c)^2+ abc( \frac{27}{a+b+c}-4)d \ \ \ \ \ \text{By AM-GM:}$

$\Leftarrow 32\ge \frac{4(a+b+c)^3}{9} (1-d)+3d(a+b+c)^2+ \frac{(a+b+c)^3}{27} \cdot( \frac{27}{a+b+c}-4)d$

$\Leftarrow 32\ge \frac{4(4-d)^3}{9} (1-d)+3d(4-d)^2+ \frac{(4-d)^3}{27} \cdot( \frac{27}{4-d}-4)d$

$(d-1)^2(6+4d-d^2)\ge 0$ . Proof of the lemma is finished.

Let $a+b+c+d=x\ge 4$ . By the lemma we have $\frac{x^4+128}{24x}\ge abc+bcd+cda+dab$ .

So it is enough to prove

$x+\frac{9\cdot 24x}{x^4+128}\ge \frac{25}{4} \ \ \ \Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 800)\ge 0$

$\Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 704)\ge 0$

$\Leftarrow \ \ (x - 4)^2 (4 x^3 + 7 x^2 - 8 x - 176)\ge 0$

Which is obvious due to $x\ge 4$ .

beautifull prove

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sqing

41825 posts

sqing

#47 h Mar 27, 2019, 3:21 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

arqady wrote:

Let $a$ , $b$ , $c$ and $d$ are positives such that $abcd=1$ . Prove that:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7$

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$

https://artofproblemsolving.com/community/c6h1810691p12061378

Let $a,b,c,d>0$ such that $abcd=1$ . prove that:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{289}{a+b+c+d+23} \ge \frac{397}{27}$

This post has been edited 3 times. Last edited by sqing, Mar 12, 2020, 2:36 AM

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Kala_Para_Na

28 posts

Kala_Para_Na

#48 h Mar 28, 2019, 5:54 PM • 1 Y

Y by Adventure10

WLOG $ab \ge cd$ $\quad$ $\therefore ab \ge 1$
$a+b \ge 2\sqrt{ab}, c+d \ge 2\sqrt{cd}=\frac{2}{\sqrt{ab}}$ , Assume $\sqrt{ab}=x$
$a+b+c+d \ge 2 (x+\frac{1}{x})$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} = \frac{a+b+c+d}{ab}+(c+d)(ab-cd)+\frac{9}{{a + b + c + d}}=\frac{k}{x^2}+\frac{9}{k}+(c+d)(x^2-\frac{1}{x^2})$ where $k=a+b+c+d$

Case 1: $3x \ge 2(x+\frac{1}{x}) \implies x \ge \sqrt{2}$

Lemma 1: $\frac{k}{x^2}+\frac{9}{k} \ge \frac{6}{x}$
Proof of lemma 1:
$\frac{k}{x^2}+\frac{9}{k} \ge \frac{6}{x} \iff k^2 + 9x^2 \ge 6xk \iff (k-3x)^2 \ge 0$ which is obviously true. $\blacksquare$

$LHS \ge \frac{6}{x}+(c+d)(x^2-\frac{1}{x^2}) \ge \frac{6}{x}+ \frac{2}{x}(x^2-\frac{1}{x^2})$

Assume $f(x)=\frac{6}{x}+ \frac{2}{x}(x^2-\frac{1}{x^2})=6x^{-1}+2x-2x^{-3}$

So, $f'(x)=\frac{2}{x^4}(x^4-3x^2+3)=\frac{2}{x^4}((x^2-1.5)^2+0.75)$ which is always positive when $x \ge \sqrt{2}$ .
which means $f(x)$ is an increasing function. So the least value can be achieved when $x = \sqrt{2}$

$\therefore LHS \ge f(\sqrt{2})=\frac{9}{\sqrt2} \ge \frac{25}{4}$

Case 2: $3x < 2(x+\frac{1}{x}) \le k$

Assume $g(k)=\frac{k}{x^2}+\frac{9}{k}$

So, $g'(k)=\frac{1}{x^2}-\frac{9}{k^2} = \frac{k^2-9x^2}{x^2k^2}$ which is positive since $k>3x$
This implies $g(k)$ is an increasing function when $k>3x$

$\therefore LHS \ge g(k)+\frac{2}{x}(x^2-\frac{1}{x^2}) \ge g(2(x+\frac{1}{x}))+\frac{2}{x}(x^2-\frac{1}{x^2})=\frac{2(x+\frac{1}{x})}{x^2}+\frac{9}{2(x+\frac{1}{x})}+\frac{2}{x}(x^2-\frac{1}{x^2})=(2x+\frac{2}{x})+\frac{9}{2x+\frac{2}{x}}=y+\frac{9}{y}$ where $y=2x+\frac{2}{x} \ge 4$

Assume $h(y)=y+\frac{9}{y}$

So, $h'(y)=1-\frac{9}{y^2}=\frac{y^2-9}{y^2}$ , since $y \ge 4 >3$ , this derivative must be positive, so $h(y)$ is an increasing function too.

Thus, $LHS \ge h(y) \ge h(4) = \frac{25}{4}$

Q.E.D

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mihaig

7346 posts

mihaig

#49 h Jun 6, 2019, 10:03 AM • 1 Y

Y by Adventure10

oldbeginner wrote:

Show us your slution, please!

If it's still needed,just contact me in PM and I'll send my solution.
At this moment,I will not post it publicly.

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Say...an

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Say...an

#50 h Jun 6, 2019, 11:21 AM • 2 Y

Y by Adventure10, Mango247

1/a+1/b+1/c+1/d>=16/a+b+c+d. a+b+c+d>=4. 1/a+1/b+1/c+1/d+ 9/a+b+c+d>=25/4

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GorgonMathDota

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GorgonMathDota

#51 h Jun 6, 2019, 11:41 AM • 2 Y

Y by Adventure10, Mango247

Say...an wrote:

1/a+1/b+1/c+1/d>=16/a+b+c+d. a+b+c+d>=4. 1/a+1/b+1/c+1/d+ 9/a+b+c+d>=25/4

How does the first one implies the second. thats totally unrelated.
And moreover, $\frac{16}{a + b + c + d} \le 4$ .

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Say...an

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Say...an

#53 h May 5, 2020, 6:14 AM

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I also wrote it...in different way ....see it carefully

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sqing

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sqing

#54 h Jun 2, 2020, 12:21 PM • 1 Y

Y by teomihai

sqing wrote:

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$

Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$ . Prove that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{\lambda }{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+\lambda }}{n}$ Where $0\le \lambda \le4(n-1).$
(SXTX (3)2020)

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sqing

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sqing

#55 h Jun 8, 2020, 12:51 PM • 1 Y

Y by Mango247

The positive reals $a,b,c,d$ satisfy $3(a^2+b^2+c^2+d^2)=4(a+b+c+d)+1$ . Prove that $a^3+b^3+c^3+d^3\leq11.$ Where?

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HamstPan38825

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#56 h Feb 28, 2025, 2:49 AM

Y by

Lagrange multipliers practice incoming.

We use Lagrange multipliers on the open set $U = \left\{a^2+b^2+c^2+d^2 < 4 \mid a, b, c, d > 4/25\right\}$ . For any point $\mathbf x \in \overline U \setminus U$ , either $U$ satisfies $a^2+b^2+c^2+d^2 = 4$ , in which case $(a, b, c, d) = (1, 1, 1, 1)$ , or $\frac 4{25} \in \{a, b, c, d\}$ , in which case the inequality is trivial. So by Lagrange multipliers, we just need to check the critical points of $f(a, b, c, d) = \frac 1a+\frac 1b+\frac 1c+\frac 1d + \frac 9{a+b+c+d}$ on the constraint set $1 = g(a, b, c, d) = abcd$ . Compute the gradients
$\begin{align*} \nabla f &= \left[-\frac 1{a^2}-\frac 9{(a+b+c+d)^2}, -\frac 1{b^2}-\frac 9{(a+b+c+d)^2}, -\frac 1{c^2}-\frac 9{(a+b+c+d)^2}, -\frac 1{d^2}-\frac 9{(a+b+c+d)^2}\right] \\ \nabla g&= [bcd, acd, abd, abc]. \end{align*}$ The critical points $\mathbf x$ are precisely those where $\nabla g = \lambda \nabla f$ for some scalar $\lambda$ . We may discard the case $\lambda=0$ and assume $\lambda \neq 0$ . These translate to the equations
$\begin{align*} bcd + \frac{\lambda}{a^2} &= acd+ \frac{\lambda}{b^2} \\ \iff cd(b-a) &= \lambda \left(\frac 1b+\frac 1a\right)\left(\frac{a-b}{ab}\right). \end{align*}$ Hence either $a=b$ or $\lambda\left(\frac 1a+\frac 1b\right) = -abcd$ and cyclic permutations. Now we split into cases:

If one pair $a = b$ , then $\frac 1b+\frac 1c=\frac 1c+\frac 1d=\frac 1d+\frac 1a$ are all equal to $\frac{-abcd}{\lambda}$ , so $a=b=c=d$ .
If $a, b, c, d$ are pairwise distinct, then $\frac 1a+\frac 1b=\frac 1b+\frac 1c=\frac 1c+\frac 1d=\frac 1d+\frac 1a$ , i.e. $a=c$ and $b=d$ .

So the critical points land where $a=b=c$ or $a=b$ and $c=d$ .

First Type: Here we may substitute for $d$ and rewrite $f(a, b, c, d) = \frac 3a+a^3+\frac{9a^3}{3a^4 + 1}$ has a global minimum equal to $\frac{25}4$ at $a=1$ , as needed.

Second Type: Here we may substitute for $c$ and write $f(a, b, c, d) = \frac 2a+2a+\frac{9a}{2a^2+2}$ has a global minimum equal to $\frac{25}4$ at $a=1$ , as needed.

Hence $f \geq \frac{25}4$ on all the critical points, and we are done.

This post has been edited 1 time. Last edited by HamstPan38825, Feb 28, 2025, 2:50 AM

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Ritwin

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#57 h Feb 28, 2025, 3:53 AM

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The inequality is clearly true if $\min(a, b, c, d) \leq 0.1$ , so it is also true if $\max(a, b, c, d) \geq 1000$ . Hence it suffices to solve the case $(a, b, c, d) \in (0.1, 1000)^4$ . This region is compact, so $f = \tfrac1a + \tfrac1b + \tfrac1c + \tfrac1d + \tfrac9{a+b+c+d}$ has a minimum, and by the previous discussion we know the minimum lies in the interior of the region.

We proceed with Lagrange multipliers. The condition is $g = abcd-1$ , where $\nabla g = \langle bcd, cda, dab, abc \rangle \propto \langle \tfrac1a, \tfrac1b, \tfrac1c, \tfrac1d \rangle.$ Letting $s = a+b+c+d$ , we also have $-f_a = \frac{1}{a^2} + \frac{9}{s^2}.$ Hence the minimum occurs somewhere along $\nabla f = \lambda' \nabla g \implies \langle a^{-2} + \kappa, b^{-2} + \kappa, c^{-2} + \kappa, d^{-2} + \kappa \rangle = \lambda \langle a^{-1}, b^{-1}, c^{-1}, d^{-1} \rangle,$ where $\kappa = 9/s^2$ . This means $a$ , $b$ , $c$ , and $d$ are each roots of $\kappa x^2 - \lambda x + 1 = 0$ , so the set $\{a, b, c, d\}$ has size at most $2$ . Also note that the product of the roots is $\kappa^{-1} = \tfrac19 s^2$ .

Now we finish by some casework.

Suppose $a=b=c=d$ . Then they're all $1$ , achieving the minimum of $\tfrac{25}{4}$ .
Suppose $a=b=c=t$ and $d=1/t^3$ with $t \neq 1$ . Then we would have $t \cdot t^{-3} = \tfrac19 \bigl(3t+t^{-3}\bigr)^2 \iff \tfrac3t = 3t + \tfrac{1}{t^3}.$ But we have $3t+t^{-3} > (2.25t^2+t^{-2})/t \geq 3/t$ , contradiction.
Suppose $a=b=t$ and $c=d=1/t$ with $t \neq 1$ . Then, letting $r = 2t+\tfrac2t \geq 4$ , we would have $f = r + \tfrac9r \geq 4 + \tfrac94 = \tfrac{25}{4},$ since $r+\tfrac9r$ is an increasing function over $r \in (3, \infty)$ .

All other cases are permutations of the above. $\blacksquare$

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