Find the largest value that the expression can take $a^3b + b^3a$ where $a, b$ are non-negative real numbers, with $a + b = 3$.
Problem
Source:
Tags: algebra, inequalities
16.09.2021 17:10
$ab(a^2+b^2)=ab(9-2ab)=9ab-2(ab)^2$, let $ab=k$, maximum of $9k-2k^2$ is when $k=\frac{9}{4}$. Plugging in gives maximum of $\frac{81}{8}$ when $a=b=\frac{3}{2}$
16.09.2021 17:26
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 $a=b=\frac{3}{2}.$
16.09.2021 17:40
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 $a=b=\frac{3}{2}.$
16.09.2021 18:40
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
16.09.2021 18:55
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 $(a,b)=\left(\frac32,\frac32\right)$.
16.09.2021 19:10
16.09.2021 19:45
16.09.2021 20:44
OlympusHero wrote:
Why did you copy my solution word for word, just with x instead of k
17.09.2021 07:01
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?
17.09.2021 08:02
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 $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$.
17.09.2021 11:15
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 $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$.
17.09.2021 11:25
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 $u=v=27-u-v \implies a=b=\frac32$
17.09.2021 14:36
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 $a=b=\frac{3}{2}.$
17.09.2021 17:10
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 $a=\frac{3\sqrt{6}-3}{5},b=\frac{18-3\sqrt{6}}{5}.$
17.09.2021 18:36
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 $2a^2b^3+ab^4 \le \frac{(x+2)^5}{xy^3} \left(\frac35\right)^5 = \frac{2916(62\sqrt6-117)}{3125}$
17.09.2021 21:06
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.
18.09.2021 00:01
Writing similar solutions is allowed. Read the C&P rules. There should be no further debate about this.
18.09.2021 02:59
18.09.2021 03:00
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 $a=b=\frac{3}{2}.$
Attachments:

18.09.2021 04:29
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 $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$
18.09.2021 23:20
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 $9b-2b^2(3-b) \le 9b \le 27$
19.09.2021 02:48
19.09.2021 08:07
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 $b=\beta=\frac{7-\sqrt{22}}{3}, a=3-\beta=\frac{2+\sqrt{22}}{3}$
19.09.2021 10:12
21.09.2021 02:58
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 $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$
Attachments:

