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

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 $a=b=\frac{3}{2}.$

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 $a=b=\frac{3}{2}.$

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

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 $(a,b)=\left(\frac32,\frac32\right)$.

OlympusHero wrote:

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

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?

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 $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$.

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 $\frac{9}{4}\left((a+b)^2-2ab\right) \rightarrow \frac{9}{4}\left(9-2\left(\frac{9}{4}\right)\right) \rightarrow \frac{9}{4}\left(9-\frac{9}{2}\right)=\boxed{\frac{81}{8}}$.

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

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 $a=b=\frac{3}{2}.$

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 $a=\frac{3\sqrt{6}-3}{5},b=\frac{18-3\sqrt{6}}{5}.$

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

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.

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

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 $a=b=\frac{3}{2}.$

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 $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$

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 $9b-2b^2(3-b) \le 9b \le 27$

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 $b=\beta=\frac{7-\sqrt{22}}{3}, a=3-\beta=\frac{2+\sqrt{22}}{3}$

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 $a=\frac{2+\sqrt{22}}{3},b=\frac{7-\sqrt{22}}{3}.$