# Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.

Let $a,b,c$ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$ab^2+bc^2+ca^2\le 4.\tag{*}$$

In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}$$
$\color{gray}{\mbox{(Without using “universal” Lagrange multipliers method).}}$

Thanks!

#### Solutions Collecting From Web of "Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$."

I prove this stronger inequality,
With loss of out,let $a=\min{(a,b,c)}$
\begin{align*}&4(a+b+c)^3-27(a^2b+b^2c+c^2a+abc)\\
&=9a(a^2+b^2+c^2-ab-bc-ac)+(4b+c-5a)(a+b-2c)^2\ge 0
\end{align*}

other nice methods:

with out loss of let $b=mid{(a,b,c)}$,then $(b-a)(b-c)\le 0$,so
$$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+c)^2=2b(a+c)(a+c)/2\le\dfrac{4}{27}(a+b+c)^3$$

Let $a=3x, b=3y, c=3z$ and $f(x,y,z)=x^2y+y^2 z +z^2 x$

Without loss of generality let’s assume that $x= \max { (x,y,z)}$

Then $f(x+\frac{z}{2}, y+\frac{z}{2}, 0) – f(x,y,z)=yz(x-y)+\frac{xz}{2}(x-z) + \frac{z^3}{8} + \frac{z^2y}{4} \ge 0$

And $f(x,y,0)= x^2y = 4 \frac{x}{2} \frac{x}{2} y \le 4\left( \frac{x+y}{3} \right)^3=\frac{4}{27}$

From AM-GM inequality.

Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.

Hence, by Rearrangement and AM-GM we obtain:
$$ab^2+bc^2+ca^2=ab\cdot b+bc\cdot c+ca\cdot a\leq xy\cdot x+xz\cdot y+yz\cdot z=y(x^2+xz+z^2)\leq$$
$$\leq y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Done!

for 2nd one:

let $a=min[a,b,c],b=a+u,c=a+v$

RHS-LHS$=(v-2u)^2(4v+u)+9av^2+18auv+27a^2v+9au^2+27a^2u+27a^3\ge0$

the “=”will hold when $a=0,v=2u\implies c=2b$.