# Kunihiko Chikaya's Inequality with a Constraint

### Solution 1

\displaystyle \begin{align} 4a^3-3a+1=(a+1)(2a-1)^2\ge 0,\\ 9b^3-3b+\frac{2}{3}=\left(b+\frac{2}{3}\right)(3b-1)^2\ge 0,\\ 36c^3-3c+\frac{1}{3}=\left(c+\frac{1}{3}\right)(6c-1)^2\ge 0. \end{align}

Summing up,

\displaystyle\begin{align} 4a^3+9b^3+36c^3&\ge 3(a+b+c)-\left(1+\frac{2}{3}+\frac{1}{3}\right)\\ &=3\cdot 1-2=1. \end{align}

### Solution 2

The required inequality is a special case of the following statement:

Assume $x+y+z=1$ and $a+b+c=1,$ $a,b,c,x,y,z \gt 0.$ Then

$\displaystyle \frac{a^3}{x^2} + \frac{b^3}{y^2} + \frac{c^3}{z^2} \ge 1.$

$\displaystyle \frac{a^3}{x^2} + \frac{b^3}{y^2} + \frac{c^3}{z^2}\ge\frac{(a+b+c)^3}{(x+y+z)^2} = 1.$

In this problem $\displaystyle x=\frac{1}{2},$ $\displaystyle y=\frac{1}{3},$ $\displaystyle z=\frac{1}{6}.$

### Solution 3

Obviously $\displaystyle \frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}+c}{6} = \frac{1}{6}.$ By the power-mean theorem

\displaystyle\begin{align} \frac{3\left(\frac{a}{3}\right)^3 +2\left(\frac{b}{2}\right)^3 + c^3}{6} &\ge\left(\frac{3\frac{a}{3}+2\frac{b}{2}+c}{6}\right)^3\\ &=\left(\frac{1}{6}\right)^3. \end{align}

Multiplying both parts by $6^3$ yields the required result.

### Solution 4

\displaystyle \begin{align} a&=\frac{4}{3}\cdot 3\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot a\le\frac{4}{3}\left(\frac{1}{8}+\frac{1}{8}+a^3\right)=\frac{1}{3}+\frac{4}{3}a^3\\ b&=3\cdot 3\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot b\le 3\left(\frac{1}{27}+\frac{1}{27}+b^3\right)=\frac{2}{9}+3b^3\\ c&=12\cdot 3\cdot \frac{1}{6}\cdot \frac{1}{6}\cdot c\le 12\left(\frac{1}{216}+\frac{1}{216}+c^3\right)=\frac{1}{9}+12c^3 \end{align}

Summing up,

$\displaystyle 1=a+b+c\le\frac{2}{3}+\frac{4}{3}a^3+3b^3+12c^3,$

i.e.,

$\displaystyle 1=3\left(1-\frac{2}{3}\right)\le 4a^3+9b^3+36c^3.$

### Solution 5

The function $v=y^3$ is convex, so that by Jensen's inequality,

\displaystyle \begin{align} 4a^3+9b^3+36c^3 &= \frac{1}{2}(2a)^3+\frac{1}{3}(3b)^3+\frac{1}{6}(6c)^3\\ &\ge\left( \frac{1}{2}(2a)+\frac{1}{3}(3b)+\frac{1}{6}(6c)\right)^3\\ &=(a+b+c)^3=1. \end{align}

### Solution 6

We can eyeball from the constraint of the Lagrangian that $3\cdot 4a^2=3\cdot 9b^2=3\cdot 36c^2,$ giving the minimum of $4a^3+9b^3+36c^3$ at $\displaystyle (a,b,c)=\left(\frac{1}{2},\frac{1}{3},\frac{1}{6}\right).$ We know it is a minimum because the second derivatives are $24a,54b, 216c$ are all positive and the Lagrangian has no cross terms.

### Acknowledgment

Kunihiko Chikaya has shared his inequality, along with a solution of his at the mathematical inequalities facebook group. Both were first posted on the web on September 13, 2014. I am grateful to Kunihiko for the permission to reproduce his post here.

Solution 2 is by Konstantin Knop; Solution 3 is by Maxim Razin; Solution 4 is by Tran Quoc Thinh; Solution 6 is by N. N. Taleb.