ab+bc+ca = 1

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Tales

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Tales

#1 h Mar 22, 2005, 8:03 PM • 8 Y

Y by itslumi, Congruentisogonal44, SmallStone9, jhu08, Adventure10, ImSh95, Mango247, and 1 other user

If $a$ , $b$ , $c$ are three positive real numbers such that $ab+bc+ca = 1$ , prove that $\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.$

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darij grinberg

6555 posts

darij grinberg

#2 h Mar 22, 2005, 9:03 PM • 16 Y

Y by ValidName, HolyMath, enzoP14, Tip_pay, ylt_chn, jhu08, ImSh95, ehuseyinyigit, adityaguharoy, Adventure10, Mango247, SmallStone9, MS_asdfgzxcvb, and 3 other users

Putting the problem into LaTeX:

Problem. Let a, b, c be three positive reals such that ab + bc + ca = 1. Prove the inequality

$\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{1}{abc}$ .

This was the 2nd problem in the 2nd German TST 2005, so you can guess what kind of list it's from

. On the exam, I messed up the problem, trying to apply all imaginable hardcore strategies and methods but failing to realize that every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution (for instance, I just couldn't believe that one could get rid of the $\sqrt[3]{...}$ roots using power means and the inequality would remain correct - we on ML are used to much sharper inequalities...). Anyway, here is the solution I found some hours after the exam, being confronted with the fact that it was just too straightforward for me:

Since ab + bc + ca = 1, we have

$\frac{1}{a}+6b=\frac{ab+bc+ca}{a}+6b=\left(b+\frac{bc}{a}+c\right)+6b=7b+c+\frac{bc}{a}$ ,

and similarly

$\frac{1}{b}+6c=7c+a+\frac{ca}{b}$ ;
$\frac{1}{c}+6a=7a+b+\frac{ab}{c}$ ,

so that

$\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)$
$=\left(7b+c+\frac{bc}{a}\right)+\left(7c+a+\frac{ca}{b}\right)+\left(7a+b+\frac{ab}{c}\right)$
$=8\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)$ .

But since

$\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)-\left(a+b+c\right)=\frac{\left(ca-ab\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2}{2abc}\geq 0$ ,

we have $a+b+c\leq \frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}$ , and thus

$8\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)=\left(6\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\right)+2\left(a+b+c\right)$
$\leq \left(6\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\right)+2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)$
$=6\left(a+b+c\right)+3\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)=3\frac{\left(ab+bc+ca\right)^2}{abc}=3\frac{1}{abc}$

(since ab + bc + ca = 1). Hence,

$\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)\leq3\frac{1}{abc}$ .

Hence,

$\frac{\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)}{3}\leq\frac{1}{abc}$ .

Now, by the power-mean inequality (more precisely, by the fact that the $\frac13$ -power mean is $\leq$ to the arithmetic mean), we have

$\left(\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\right)^3\leq\frac{\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)}{3}$ ,

so that

$\left(\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\right)^3\leq\frac{1}{abc}$ .

Taking the cubic root of this, we get

$\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\leq\frac{1}{\sqrt[3]{abc}}$ ,

so that

$\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{3}{\sqrt[3]{abc}}$ .

Now, by the AM-GM inequality, applied to the numbers ab, bc and ca, we have $\sqrt[3]{ab\cdot bc\cdot ca}\leq\frac{ab+bc+ca}{3}$ . Since $ab\cdot bc\cdot ca=\left(abc\right)^2$ and ab + bc + ca = 1, this becomes $\sqrt[3]{\left(abc\right)^2}\leq\frac13$ , so that $3\sqrt[3]{\left(abc\right)^2}\leq 1$ .

Thus,

$\frac{3}{\sqrt[3]{abc}}=\frac{3\sqrt[3]{\left(abc\right)^2}}{\sqrt[3]{\left(abc\right)^3}}=\frac{3\sqrt[3]{\left(abc\right)^2}}{abc}\leq\frac{1}{abc}$ .

Altogether,

$\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{3}{\sqrt[3]{abc}}\leq\frac{1}{abc}$ .

And the problem is solved.

Darij

This post has been edited 2 times. Last edited by darij grinberg, Jan 26, 2006, 10:11 PM

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namanhams

46 posts

namanhams

#3 h Mar 26, 2005, 9:27 AM • 5 Y

Y by Adventure10, jhu08, ImSh95, Infinityfun, Mango247

Ok ,here is my solution :

Applying Bunhiacopski inequality we have :

( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3:

so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3:
Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27
Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

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Loser

70 posts

Loser

#4 h May 2, 2005, 11:58 AM • 4 Y

Y by Adventure10, Adventure10, ImSh95, Mango247

namanhams wrote:

Ok ,here is my solution :

Applying Bunhiacopski inequality we have :

( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3:

so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3:
Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27
Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers.
:-?

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Soarer

2589 posts

Soarer

#5 h May 2, 2005, 12:07 PM • 3 Y

Y by Adventure10, ImSh95, Mango247

Loser wrote:

namanhams wrote:

Ok ,here is my solution :

Applying Bunhiacopski inequality we have :

( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3:

so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3:
Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27
Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers.
:-?

that's holder.

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Loser

70 posts

Loser

#6 h May 3, 2005, 5:43 PM • 3 Y

Y by Adventure10, ImSh95, Mango247

siuhochung wrote:

Loser wrote:

namanhams wrote:

Ok ,here is my solution :

Applying Bunhiacopski inequality we have :

( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3:

so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3:
Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27
Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers.
:-?

that's holder.

I don't get it again?
What do you mean by "holder"?

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Soarer

2589 posts

Soarer

#7 h May 4, 2005, 11:22 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

Holder inequality.
You can find much information about it in Internet.

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Loser

70 posts

Loser

#8 h May 4, 2005, 5:22 PM • 4 Y

Y by Adventure10, Adventure10, ImSh95, Mango247

Ok,10x very much for helping me.

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indybar

398 posts

indybar

#9 h Jul 22, 2005, 5:03 PM • 2 Y

Y by Adventure10, ImSh95

darij grinberg wrote:

every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution

How to use Jensen here? Can you show?

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shyong

1677 posts

shyong

#10 h Jul 24, 2005, 11:25 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

namanhams wrote:

Ok ,here is my solution :

Applying Bunhiacopski inequality we have :

( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3:

so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3:
Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27
Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

should be equivalent to $(abc)^2 \leq \frac{1}{27}$ ....

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k2c901_1

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k2c901_1

#11 h Aug 12, 2005, 6:23 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

Incidentally, this is also the 4th problem of the 1st TST of Taiwan in 2005.

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idioteque

75 posts

idioteque

#12 h Aug 20, 2005, 1:11 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

I wrote sth really stupid

in here and it looks like forum has some problems and I can't delete post..

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Mildorf

785 posts

Mildorf

#13 h Aug 20, 2005, 7:00 PM • 3 Y

Y by Adventure10, ImSh95, Mango247

indybar wrote:

darij grinberg wrote:

every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution

How to use Jensen here? Can you show?

That was my solution when it appeared in a WOP exam. Apply Jensen to the concave function $f(x) = \sqrt[3]{x}$ . You get a sharper inequality that can be easily Muirhead bashed after homogenizing and the substitution $x = bc, y = ca, z = ab$ .

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Chang Woo-JIn

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Chang Woo-JIn

#14 h Sep 23, 2005, 6:50 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

Here's my solution

by Jensen Ineq on $f(x)= x^{\frac{1}{3}}$ .

we get, $\sum_{cy} (\frac{1}{a}+ 6b)^ {\frac{1}{3}} \leq 3 ( \frac{1}{3} ( \sum_{cy} \frac{1}{a} + 6\sum_{cy} a ) )^{\frac{1}{3}}$ .

So, It is enough to prove that

$3 ( \frac{1}{3} ( \sum_{cy} \frac{1}{a} + 6\sum_{cy} a ) )^{\frac{1}{3}}\leq \frac{1}{abc}$ .

$\Leftrightarrow 9 \sum_{cy} a^3 b^3 c^2 + 54\sum_{cy} a^4 b^3 c^3 \leq 1$

Let $ab=p, bc=q, ca=r$ . then $p+q+r=1$

$\Leftrightarrow 9(\sum p^2 qr)(p+q+r) + 54\sum_{cy} p^2q^2 r \leq (p+q+r)^5$

$\Leftrightarrow \sum_{cy} p^5 + 5 \sum_{sy}p^4 q+ 10\sum_{sy}p^3q^2 +11\sum_{cy}p^3qr \geq 42\sum_{cy}p^2q^2 r$ .

by Muirhead Ineq, This is true.

$Q.E.D.$

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Singular

749 posts

Singular

#15 h Sep 23, 2005, 6:55 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

$(\frac{1}{a} + 3b + 3b)^{1/3} \le 2b + \frac{1}{3a}$

It becomes $3\sum a \le 1/abc$ or $3\sum a \le \sum (ab+bc+ca)/a$ .

This becomes $\sum bc/a \ge \sum a$ . By Muirhead its true.

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ehuseyinyigit

769 posts

ehuseyinyigit

#108 h Dec 4, 2023, 2:34 PM

Y by

Generalization 1
Let $a,b,c,\theta,\lambda$ be positive reals such that $ab+bc+ca=\beta$ and
$\beta ^4\left(3\theta +\lambda \beta\right)=3^{5-n}$ equalities hold. Then prove that

$\sqrt[n]{\dfrac{\theta}{a}+\lambda b}+\sqrt[n]{\dfrac{\theta}{b}+\lambda c}+\sqrt[n]{\dfrac{\theta}{c}+\lambda a}\leq \dfrac{1}{\sqrt[3]{\left(abc\right)^n}}$

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OronSH

1714 posts

OronSH

#109 h Jan 6, 2024, 2:52 AM • 4 Y

Y by teomihai, megarnie, EpicBird08, pog

By AM-GM $\frac 1a+3b+\left(\frac1{3b}+2a\right)\ge 3\sqrt[3]{\frac 1a+6b}$ so $\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \le \sum_{\text{cyc}}\left(\frac4{9a}+\frac{5a}3\right).$ Now we have $\frac 1{abc}=\frac1a+\frac1b+\frac1c$ so we want $\frac53(a+b+c)\le\frac59\left(\frac1a+\frac1b+\frac1c\right)$ which follows by Muirhead after homogenizing and simplifying.

This post has been edited 1 time. Last edited by OronSH, Jan 6, 2024, 2:56 AM

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Scilyse

385 posts

Scilyse

#110 h Feb 8, 2024, 11:04 AM

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Note that $\sum \sqrt[3]{\frac{1}{a} + 6b} \stackrel{\text{PM}}{\leq} 3 \sqrt[3]{\frac{\sum \left(\frac{1}{a} + 6a\right)}{3}}\text{.}$ Thus it suffices to prove that
$\begin{align*} 3 \sqrt[3]{\frac{\sum \left(\frac{1}{a} + 6a\right)}{3}} &\leq \frac{1}{abc} \\ \sum \left(\frac{1}{a} + 6a\right) &\leq \frac{1}{9} \cdot \left(\frac{1}{abc}\right)^3 \\ \sum \left(9bc + 54a^2 bc\right) &\leq \left(\frac{1}{abc}\right)^2 \\ 9 + 54 \sum a^2 bc &\leq \frac{1}{a^2 b^2 c^2}\text{.} \end{align*}$ Now substitute $x = bc, y = ca, z = ab$ so we get $9 + 54 \sum xy \leq \frac{1}{xyz}\text{.}$ Note that $x + y + z = 1$ .

However, $2 \sum xy = \left(\sum x\right)^2 - \sum x^2 = 1 - \sum x^2 \stackrel{\text{PM}}{\leq} \frac{2}{3}$ so therefore $9 + 54 \sum xy \leq 27\text{.}$ Additionally, $27 \leq \frac{1}{xyz}$ by AM-GM so we are done.

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kuzeyaloglu

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kuzeyaloglu

#111 h Feb 23, 2024, 8:40 PM

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By Hölder's Inequality we have $((\sum ab)^\frac{1}{3})((\sum \frac{1}{b})^ \frac{1}{3})((\sum 6ab+1)^\frac{1}{3}) \ge ({(\sum \frac{6ab+1}{a}) ^ \frac{1}{3}) }$

We need to show $\frac{1}{abc} \ge ((\sum ab)^\frac{1}{3})((\sum \frac{1}{b})^ \frac{1}{3})((\sum 6ab+1)^\frac{1}{3}) = \frac{9^ \frac{1}{3}}{(abc)^ \frac{1}{3}} \iff \frac{1}{3} \ge abc$ which is true by AM-GM on $ab,ac,bc$

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blueprimes

298 posts

blueprimes

#112 h Jun 17, 2024, 2:31 AM • 1 Y

Y by teomihai

By Holder, we have
$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{ \frac{1}{b} + 6c} + \sqrt[3]{ \frac{1}{c} + 6a} \le \left[ \left( \sum_{\text{cyc}} (6ab + 1) \right) \left( \sum_{\text{cyc}} \frac{1}{a} \right) \left( 1 + 1 + 1 \right) \right]^{1/3} = \frac{3}{(abc)^{1/3}}$ so it is sufficient to show
$\frac{3}{(abc)^{1/3}} \le \frac{1}{abc} \iff (abc)^2 \le \frac{1}{27} = \left( \frac{ab + bc + ca}{3} \right)^3$ which is true by AM-GM.

This post has been edited 1 time. Last edited by blueprimes, Jun 17, 2024, 2:31 AM

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crezk

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crezk

#113 h Jul 30, 2024, 1:50 PM • 3 Y

Y by teomihai, ehuseyinyigit, OronSH

$(a+\frac{1}{6b})^\frac{1}{3}\leq \frac{a+b}{3ab(1-ab)}-ab+\frac{1}{3} \implies (a+\frac{1}{6b})^\frac{1}{3}\leq \frac{1}{3abc}-ab +\frac{1}{3}$
then $\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}- (ab+ac+bc)+1=\frac{1}{abc}$

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numbertheory97

39 posts

numbertheory97

#114 h Aug 24, 2024, 12:00 AM

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We use Hölder. Notice that $\left(\sum_\text{cyc} \sqrt[3]{\frac1a + 6b}\right)^3 = \left(\sum_\text{cyc} \sqrt[3]{\frac{6ab + 1}{a}}\right)^3$ $\leq 3\left(\frac1a + \frac1b + \frac1c\right)\left(\sum_\text{cyc} (6ab + 1)\right) = 27\left(\frac1a + \frac1b + \frac1c\right),$ and so it suffices to show that $27\left(\frac1a + \frac1b + \frac1c\right) \leq \frac{1}{(abc)^3}.$ But upon multiplying both sides by $abc$ and rearranging this becomes $abc \leq \frac{1}{3\sqrt3}$ , which follows since $\frac13 = \frac{ab + bc + ca}{3} \geq (abc)^\frac23$ by AM-GM. $\square$

I originally found a solution involving Jensen and a lot of annoying expansion, but I switched to this one after happily discovering that Hölder works.

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little-fermat

147 posts

little-fermat

#115 h Sep 27, 2024, 3:12 AM • 1 Y

Y by teomihai

I have discussed this problem as an application of Holder in my inequality YouTube tutorial on my channel "little fermat". Here is the Video

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teomihai

2971 posts

teomihai

#116 h Sep 27, 2024, 3:20 AM • 1 Y

Y by little-fermat

very nice solution- video!

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smileapple

1007 posts

smileapple

#117 h Sep 28, 2024, 7:44 AM • 1 Y

Y by teomihai

By Hölder we find that $\left(\sum_\text{cyc}\sqrt[3]{\frac1a+6b}\right)^3\left(\sum_\text{cyc}(1+6ab)^3a\right)\le\left(\sum_\text{cyc}(1+6ab)\right)^4=9^4.$ This is enough to finish upon observing that
$\begin{align*} \sum_\text{cyc}(1+6ab)^3a&=\sum_\text{cyc}(7a+bc+ca)^3(ab+bc+ca)a\\ &\ge\sum_\text{cyc}(9a^\frac89b^\frac89c^\frac89)^3(3a^\frac23b^\frac23c^\frac23)a=3\cdot9^3\sum_\text{cyc}a^\frac{13}3b^\frac{10}3c^\frac43\\ &\ge9^4a^3b^3c^3\end{align*}$ due to the usual AM-GM. $\blacksquare$

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bin_sherlo

661 posts

bin_sherlo

#118 h Oct 8, 2024, 8:29 PM • 1 Y

Y by OronSH

$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \overset{?}{\leq} \frac{1}{abc}.$ Let $bc=\frac{x^2}{3},ac=\frac{y^2}{3},ab=\frac{z^2}{3}\iff a=\frac{yz}{x\sqrt3},b=\frac{xz}{y\sqrt3},c=\frac{xy}{z\sqrt3}$ . We have the condition $x^2+y^2+z^2=3$ .
$\sum{\sqrt[3]{\frac{x\sqrt{3}}{yz}+\frac{2\sqrt{3}xz}{y}}}\overset{?}{\leq} \frac{3\sqrt3}{xyz}$ $\sqrt[3]{\frac{x\sqrt3}{yz}+\frac{2\sqrt3xz}{y}}=\sqrt[3]{\frac{x\sqrt3}{y}(\frac{1}{z}+2z)}\leq \frac{\frac{x\sqrt3}{y}+(\frac{1}{\sqrt3z}+\frac{2z}{\sqrt3})+\sqrt3}{3}$ Thus, we conclude that
$\sum{\sqrt[3]{\frac{x\sqrt3}{yz}+\frac{2\sqrt3xz}{y}}}\leq \sqrt{3}\sum{(\frac{z}{3y}+\frac{1}{3}+\frac{1}{9x}+\frac{2x}{9})}\overset{?}{\leq} \frac{3\sqrt3}{xyz}$ $1+\frac{1}{3}\sum{\frac{z}{y}}+\frac{1}{9}\sum{\frac{1}{x}}+\frac{2}{9}\sum{x}\overset{?}{\leq }\frac{3}{xyz}$ $27\overset{?}{\geq} 2xyz\sum{x}+\sum{xy}+3\sum{x^2y}+9xyz$ We have $\sum{x}\leq \sqrt{3\sum{x^2}}=3$ and $xyz\leq 1$ thus, $2xyz\sum{x}\leq 6$ and $\sum{xy}\leq \sum{x^2}=3$ .
$6\overset{?}{\geq} \sum{x^2y}+3xyz$ By Schur with degree $4$ , we have
$9-2\sum{x^2y^2}+\sum{x^2yz}=\sum{x^4}+\sum{x^2yz}\geq \sum_{sym}{x^3y}=\sum{xy(x^2+y^2)}\geq 2\sum{x^2y^2}$ $9+\sum{x^2yz}\geq 4\sum{x^2y^2}\geq 8\sum{x^2y}-4\sum{x^2}=8\sum{x^2y}-12\implies \frac{21+\sum{x^2yz}}{8}\geq \sum{x^2y}$ Hence
$\sum{x^2y}+3xyz\leq \frac{21+\sum{x^2yz}}{8}+3xyz\leq 3+3=6$ As desired. $\blacksquare$

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onyqz

195 posts

onyqz

#119 h Nov 6, 2024, 8:25 PM • 1 Y

Y by teomihai

solution for storage

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Maximilian113

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Maximilian113

#120 h Dec 3, 2024, 3:37 PM

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Note that by Holder's Inequality, $\text{LHS} \leq (1+1+1)^{1/3} \left(\sum_{cyc} (6ab+1) \right)^{1/3}\left(\sum_{cyc}\frac{1}{a} \right)^{1/3} = \frac{3}{(abc)^{1/3}}.$ Thus it suffices to show that $3(abc)^{2/3} \leq 1 = ab+bc+ca$ but this holds by AM-GM. QED

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InftyByond

193 posts

InftyByond

#121 h Today at 1:39 AM • 1 Y

Y by Maximilian113

skibidi sigma, rizztastic ohiomaxxing problem
considering I was able to solve this problem despite applying 3 inequalities makes it incredibly weak
also WATTESIGMA IS THE HOLDER SOLVE im too bad for this
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cj13609517288

1855 posts

cj13609517288

#122 h Today at 1:03 PM • 1 Y

Y by OronSH

Note that
$\sqrt[3]{\frac1a+6b}=\sqrt[3]{\left(\frac1a\right)\left(\frac{1}{\sqrt3}+2\sqrt3 ab\right)(\sqrt3)}\le\frac{\frac1a+\frac{1}{\sqrt3}+2\sqrt3 ab+\sqrt3}{3}.$ The cyclic sum of this is
$\frac{\frac1a+\frac1b+\frac1c+4\sqrt3+2\sqrt3}{3}=\frac{1}{3abc}+2\sqrt3.$ Therefore, it suffices to prove that $abc\le\frac{1}{3\sqrt3}$ , which is obvious. $\blacksquare$

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