i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$

http://www.kalva.demon.co.uk/canada/casoln/csol995.html here?

// there was attempt to prove that $a^2b + b^2c + c^2a \le \frac{27}{8}$ but my proof was wrong I wonder if it's true

You would also be able to use Lagrange Multipliers, since we have the additional condition that $a+b+c = 3$.

N.T.TUAN wrote: Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$. Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality. After homogenization, it's equivalent to: $\sum a^3+3\sum a^2b+6abc \ge 5\sum ab^2$ $\Leftrightarrow (a+b-c)(a-b)^2 + (4a-4b+c)(a-c)(b-c) + 3abc \ge 0$ What is obviously true by condition. Equality holds when $a=b=\frac{3}{2}$ and $c=0$. $\blacksquare$

gollywog wrote: i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$ Solution. Let $f(x;y;z)=x^2y+y^2z+z^2x.$ WE wish to determine where $f$ is maximal. Since $f$ is cyclic Wlog we may assume that $x=max(x;y;z).$ Since $f(x;y;z)-f(x;z;y)=x^2y+y^2z+z^2x-x^2z-z^2y-y^2x=(y-z)(x-z)(x-y)$ we may also assume $y\geq z.$ Then $f(x+z;y;0)-f(x;y;z)=(x+z)^2y-x^2y-y^2z-z^2x=z^2x+yz(x-y)+xz(y-z)\geq 0$ So we may now assume z=0 The rest follows from AM-GM inequality $f(x;y;0)=\frac{2x^2y}{2}\leq \frac{1}{2}({\frac{x+x+2y}{3})^3=\frac{4}{27}.}$ Equality occurs when $x=2y$, hence when $(x;y;z)$ equals $(\frac{2}{3};\frac{1}{3};0), (\frac{1}{3};0;\frac{2}{3})$ or $(0;\frac{2}{3};\frac{1}{3})$.

I think the inequality is quite weak. Well I substituted (a+b+c)^3 = 27 and then applied the Schurhead inequality to get the answer.

For #1 With conditions $a\ge{b}\ge{c}\ge{0}$ and $a+b+c=3$,we have \[\frac{27}{8}-(ab^2+bc^2+ca^2)=\frac{1}{2}(a-b)(a-c)(b-c)+\frac{1}{8}(a+b-c)(a-b)^2+\frac{1}{8}c(a-c)(b-c)+\frac{3}{8}abc\ge{0}\]

vedantbonde wrote: I think the inequality is quite weak. Well I substituted (a+b+c)^3 = 27 and then applied the Schurhead inequality to get the answer. And I always thought Schur/Muirhead only work for symmetric inequalities while this one is cyclic... Looking forward to see your full solution!

N.T.TUAN wrote: Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$. Prove that $ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$ and find cases of equality. The given inequality is equivalent to, $$\frac{27}{8} \geq ab^2+bc^2+ca^2 \Leftrightarrow (a+b+c)^3 \geq 8(ab^2+bc^2+ca^2)$$ $$\Leftrightarrow a^3+b^3+c^3+6abc + 3(ab^2+bc^2+ca^2) +3(a^2b+b^2c+c^2a) \geq 8(ab^2+bc^2+ca^2)$$ Using schur's inequality, we have $$a^3+b^3+c^3+3abc \geq (ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a)$$ Therefore, it suffices to show that $$(ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a) + 3(ab^2+bc^2+ca^2) +3(a^2b+b^2c+c^2a) \geq 8(ab^2+bc^2+ca^2)$$ $$\Leftrightarrow 4(a^2b+b^2c+c^2a) - 4(ab^2+bc^2+ca^2) \geq 0$$ $$\Leftrightarrow 4(a-b)(b-c)(c-a) \geq 0$$ The last inequality is true since we are given $a \geq b \geq c$ Hence, the desired inequality is also true. Equality occues iff $(a,b,c) =(1,0,0), (0,1,0) , (0,0,1)$.

gollywog wrote: i remember similar problem (but definately not the same and it's likely that solutions is completely different). the difference was that $a+b+c=1$ and $LHS \leq 4/27$ https://artofproblemsolving.com/community/c6h497887p2797425 N.T.TUAN wrote: Let $a\geq b\geq c\geq 0$ are real numbers such that $a+b+c=3$. Prove that $$ab^{2}+bc^{2}+ca^{2}\leq\frac{27}{8}$$and find cases of equality.

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