max of a^3b + b^3a if a,b>=0, a+b=3 Puerto Rico TST 2019.3

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parmenides51

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parmenides51

#1 h Sep 16, 2021, 2:03 PM • 3 Y

Y by HWenslawski, chessgocube, jhu08

Find the largest value that the expression can take $a^3b + b^3a$ where $a, b$ are non-negative real numbers, with $a + b = 3$ .

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Arrowhead575

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Arrowhead575

#2 h Sep 16, 2021, 2:10 PM • 3 Y

Y by HWenslawski, chessgocube, NealShrestha

$ab(a^2+b^2)=ab(9-2ab)=9ab-2(ab)^2$ , let $ab=k$ , maximum of $9k-2k^2$ is when $k=\frac{9}{4}$ . Plugging in gives maximum of $\frac{81}{8}$ when $a=b=\frac{3}{2}$

This post has been edited 1 time. Last edited by Arrowhead575, Sep 16, 2021, 2:10 PM

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sqing

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sqing

#3 h Sep 16, 2021, 2:26 PM • 2 Y

Y by chessgocube, jhu08

parmenides51 wrote:

Find the largest value that the expression can take $a^3b + b^3a$ where $a, b$ are non-negative real numbers, with $a + b = 3$ .

$a^3b + b^3a=\frac{1}{2}\cdot 2ab(9-2ab)\leq \frac{1}{2}\cdot\left(\frac{ 2ab+9-2ab}{2}\right)^2=\frac{81}{8}$ Equality holds when $a=b=\frac{3}{2}.$

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sqing

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sqing

#4 h Sep 16, 2021, 2:40 PM • 2 Y

Y by chessgocube, jhu08

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$3a^3b^2+ab^4\leq\frac{243}{8}$ Equality holds when $a=b=\frac{3}{2}.$

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e61442289

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e61442289

#5 h Sep 16, 2021, 3:40 PM • 2 Y

Y by chessgocube, jhu08

parmenides51 wrote:

Find the largest value that the expression can take $a^3b + b^3a$ where $a, b$ are non-negative real numbers, with $a + b = 3$ .

note that a³b + b³a = ab (a² + b²)
also (a² + b²) = (a + b) ²-2ab = 9-2ab
then by MA - MG we have ab ≤ 9/4
then ab (9-2ab) ≤ 81/4
and the equality is obtained with a = b = 3/2

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jasperE3

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jasperE3

#6 h Sep 16, 2021, 3:55 PM • 2 Y

Y by chessgocube, jhu08

Let $t=ab$ :
$a^3b+b^3a=ab((a+b)^2-2ab)=\frac12\cdot2t(9-2t)\le\frac12\left(\frac92\right)^2=\frac{81}8,$ equality when $(a,b)=\left(\frac32,\frac32\right)$ .

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OlympusHero

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OlympusHero

#7 h Sep 16, 2021, 4:10 PM • 2 Y

Y by chessgocube, jhu08

Solution

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kante314

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kante314

#8 h Sep 16, 2021, 4:45 PM • 2 Y

Y by chessgocube, jhu08

Solution

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Arrowhead575

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Arrowhead575

#9 h Sep 16, 2021, 5:44 PM

Y by

OlympusHero wrote:

Solution

Why did you copy my solution word for word, just with x instead of k

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OlympusHero

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OlympusHero

#10 h Sep 17, 2021, 4:01 AM • 1 Y

Y by jhu08

I did not copy your solution word for word. Writing similar solutions is allowed. Please read the rules before making comments like this. Also, literally everyone used the exact same method, why are you picking on me specifically?

This post has been edited 1 time. Last edited by OlympusHero, Sep 17, 2021, 4:01 AM

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Bowser498

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Bowser498

#11 h Sep 17, 2021, 5:02 AM • 2 Y

Y by jhu08, oralayhan

The given condition implies $a^2+2ab+b^2=9$ , and by AM-GM: $\frac{a^3b+b^3a}{2} \geq \sqrt{(ab)^4} \rightarrow \frac{ab(a^2+b^2)}{2} \geq (ab)^2 \rightarrow ab(a^2+b^2) \geq 2(ab)^2 \rightarrow a^2+b^2 \geq 2ab$ . Rewrite as $9-2ab \geq 2ab \rightarrow 9 \geq 4ab \rightarrow ab \leq \frac{9}{4}$ . To maximize the value of the given expression, use $ab=\frac{9}{4}$ to equivalently force us to find the maximum of $\frac{9}{4}(a^2+b^2)$ . But $a^2+b^2=(a+b)^2-2ab$ , so we have $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$ .

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MihaiT

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MihaiT

#12 h Sep 17, 2021, 8:15 AM • 2 Y

Y by oralayhan, jhu08

splendid!!

Bowser498 wrote:

The given condition implies $a^2+2ab+b^2=9$ , and by AM-GM: $\frac{a^3b+b^3a}{2} \geq \sqrt{(ab)^4} \rightarrow \frac{ab(a^2+b^2)}{2} \geq (ab)^2 \rightarrow ab(a^2+b^2) \geq 2(ab)^2 \rightarrow a^2+b^2 \geq 2ab$ . Rewrite as $9-2ab \geq 2ab \rightarrow 9 \geq 4ab \rightarrow ab \leq \frac{9}{4}$ . To maximize the value of the given expression, use $ab=\frac{9}{4}$ to equivalently force us to find the maximum of $\frac{9}{4}(a^2+b^2)$ . But $a^2+b^2=(a+b)^2-2ab$ , so we have $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$ .

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DAVROS

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DAVROS

#13 h Sep 17, 2021, 8:25 AM • 1 Y

Y by jhu08

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$3a^3b^2+ab^4\leq\frac{243}{8}$

$LHS = ab^2(3a^2+b^2) = ab^2((a+b)(3a+b)-4ab) = ab^2(27-6b-4ab)$

Put $u=6b, v=4ab$ then this is $\frac{1}{24}uv(27-u-v) \le \frac{1}{24}\left(\frac{u+v+27-u-v}{3}\right)^3=\frac{243}{8}$

Equality when $u=v=27-u-v \implies a=b=\frac32$

This post has been edited 1 time. Last edited by DAVROS, Sep 17, 2021, 8:31 AM

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sqing

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sqing

#14 h Sep 17, 2021, 11:36 AM • 1 Y

Y by jhu08

Nice.Thanks.

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$3a^3b^2+ab^4\leq\frac{243}{8}$ Equality holds when $a=b=\frac{3}{2}.$

Solution of DAVROS:
$3a^3b^2+ab^4= ab^2((a+b)(3a+b)-4ab) = ab^2(27-6b-4ab)$ $=\frac{1}{24} 6b\cdot 4ab(27- 6b-4ab) \leq \frac{1}{24}\left(\frac{6b+4ab+27-6b-4ab}{3}\right)^3=\frac{243}{8}$ Equality holds when $a=b=\frac{3}{2}.$

This post has been edited 2 times. Last edited by sqing, Sep 17, 2021, 11:45 AM

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sqing

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sqing

#15 h Sep 17, 2021, 2:10 PM • 1 Y

Y by jhu08

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$2a^2b^3+ab^4 \leq \frac{2916(62\sqrt{6}-117)}{3125}$ Equality holds when $a=\frac{3\sqrt{6}-3}{5},b=\frac{18-3\sqrt{6}}{5}.$

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DAVROS

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DAVROS

#16 h Sep 17, 2021, 3:36 PM • 1 Y

Y by jhu08

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$2a^2b^3+ab^4 \leq \frac{2916(62\sqrt{6}-117)}{3125}$ Equality holds when $a=\frac{3\sqrt{6}-3}{5},b=\frac{18-3\sqrt{6}}{5}.$

By AM-GM : $2a^2b^3+ab^4 = \frac{1}{xy^3} (xa)(yb)^3(2a+b) \le \frac{1}{xy^3} \left(\frac{xa+3yb+2a+b}{5}\right)^5$

For a useful implementation you require $x+2=3y+1, xa=yb=2a+b, a+b=3$ for which the all-positive solution is

$a=\frac{3\sqrt6-3}{5}, b=\frac{18-3\sqrt6}{5},x=2+\sqrt6, y=1+\sqrt{\frac23}$ where this $a,b$ are values at equality

Sub in inequality for $2a^2b^3+ab^4 \le \frac{(x+2)^5}{xy^3} \left(\frac35\right)^5 = \frac{2916(62\sqrt6-117)}{3125}$

This post has been edited 2 times. Last edited by DAVROS, Sep 17, 2021, 3:44 PM
Reason: wrong quote

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Arrowhead575

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Arrowhead575

#17 h Sep 17, 2021, 6:06 PM

Y by

OlympusHero wrote:

I did not copy your solution word for word. Writing similar solutions is allowed. Please read the rules before making comments like this. Also, literally everyone used the exact same method, why are you picking on me specifically?

It seems you did the same exact thing with x instead of k. Solutions should give a new insight on a problem, not just to brag that you know how to solve it. Also, I don't mean to pick on you, but you have been doing the same thing with not only my posts but also others.
@below it’s not similar it is word for word the exact same, with a different variable. You’re right, there should be no argument about this.

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OlympusHero

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OlympusHero

#18 h Sep 17, 2021, 9:01 PM • 1 Y

Y by jhu08

Writing similar solutions is allowed. Read the C&P rules. There should be no further debate about this.

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sqing

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sqing

#19 h Sep 17, 2021, 11:59 PM • 1 Y

Y by jhu08

Very nice.Thank DAVROS.

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sqing

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sqing

#20 h Sep 18, 2021, 12:00 AM • 1 Y

Y by jhu08

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$3a^3b^2+ab^4\leq\frac{243}{8}$ Equality holds when $a=b=\frac{3}{2}.$

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sqing

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#21 h Sep 18, 2021, 1:29 AM • 1 Y

Y by jhu08

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$a^2b+b^3\leq27$ Equality holds when $a=0,b=3.$
$a^2b+b\leq\frac{4(9+\sqrt{6})}{9}$ Equality holds when $a=1+\sqrt{\frac{2}{3}},b=2-\sqrt{\frac{2}{3}}.$
$2a^2b-b\leq 2+\frac{7\sqrt{42}}{9}$ Equality holds when $a=1+\sqrt{\frac{7}{6}},b=2-\sqrt{\frac{7}{6}}.$
$3a^2b+b\leq 8+\frac{32\sqrt{2}}{9}$ Equality holds when $a=1+\frac{2\sqrt{2}}{3},b=2-\frac{2\sqrt{2}}{3}.$
$2a^2b+b^2\leq\frac{451+13\sqrt{13}}{54}$ Equality holds when $a=\frac{7+\sqrt{13}}{6},b=\frac{11-\sqrt{13}}{6}.$
$3a^2b+b^2\leq\frac{2567+92\sqrt{46}}{243}$ Equality holds when $a=\frac{10+\sqrt{46}}{9},b=\frac{17-\sqrt{46}}{9}.$
$a^2b-b^2\leq\frac{44\sqrt{22}-119}{27}$ Equality holds when $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$

This post has been edited 1 time. Last edited by sqing, Sep 18, 2021, 1:42 AM

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DAVROS

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DAVROS

#22 h Sep 18, 2021, 8:20 PM

Y by

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$a^2b+b^3\leq27$ Equality holds when $a=0,b=3.$

Sub $a=3-b$ so for $b\in[0,3]$ you have $9b-2b^2(3-b) \le 9b \le 27$

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sqing

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#23 h Sep 18, 2021, 11:48 PM

Y by

Thank DAVROS.

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DAVROS

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DAVROS

#24 h Sep 19, 2021, 5:07 AM

Y by

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that

$a^2b-b^2\leq\frac{44\sqrt{22}-119}{27}$ Equality holds when $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$

Sub $a=3-b$ then you want $\max {(b^3-7b^2+9b)} \text{ over } [0,3]$

You can write $b^3-7b^2+9b = \alpha\beta^2+(b-\alpha)(b-\beta)^2$ where $\alpha+2\beta=7\text{ and }\beta^2+2\alpha\beta=9$

Using the solution $\alpha = \frac{7+2\sqrt{22}}{3}, \beta=\frac{7-\sqrt{22}}{3}$ and noting that $b-\alpha <0$ over $[0,3]$ this gives

$b^3-7b^2+9b \le \alpha\beta^2 = \frac{44\sqrt{22}-119}{27}$ with equality at $b=\beta=\frac{7-\sqrt{22}}{3}, a=3-\beta=\frac{2+\sqrt{22}}{3}$

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sqing

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sqing

#25 h Sep 19, 2021, 7:12 AM

Y by

Nice. Thank DAVROS.

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sqing

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sqing

#26 h Sep 20, 2021, 11:58 PM

Y by

sqing wrote:

Let $a, b$ are non-negative real numbers such that $a + b = 3.$ Prove that
$a^2b+b^3\leq27$ Equality holds when $a=0,b=3.$
$a^2b+b\leq\frac{4(9+\sqrt{6})}{9}$ Equality holds when $a=1+\sqrt{\frac{2}{3}},b=2-\sqrt{\frac{2}{3}}.$
$2a^2b-b\leq 2+\frac{7\sqrt{42}}{9}$ Equality holds when $a=1+\sqrt{\frac{7}{6}},b=2-\sqrt{\frac{7}{6}}.$
$3a^2b+b\leq 8+\frac{32\sqrt{2}}{9}$ Equality holds when $a=1+\frac{2\sqrt{2}}{3},b=2-\frac{2\sqrt{2}}{3}.$
$2a^2b+b^2\leq\frac{451+13\sqrt{13}}{54}$ Equality holds when $a=\frac{7+\sqrt{13}}{6},b=\frac{11-\sqrt{13}}{6}.$
$3a^2b+b^2\leq\frac{2567+92\sqrt{46}}{243}$ Equality holds when $a=\frac{10+\sqrt{46}}{9},b=\frac{17-\sqrt{46}}{9}.$
$a^2b-b^2\leq\frac{44\sqrt{22}-119}{27}$ Equality holds when $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$

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