prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$

Question

prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .

I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!

Answer 1

$$\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}=\frac{a^2}{6}+\frac{a^2}{6}+\frac{a^2}{6}+\frac{b^3}{6}+\frac{b^3}{6}+\frac{c^6}{6}.$$

Then use AM-GM with these $6$ variables.

Answer 2

A slightly different approach using $\frac{a^2}2+\frac{b^2}2\ge ab$ : $$\begin{align*} \frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}+\frac{abc}3&=\frac13 a^2+\frac13 abc+\left(\frac16 a^2+ \frac16 b^3\right)+\left(\frac16 b^3+\frac16 c^6\right)\\&\ge\frac13a^2+\frac13abc+\frac13ab^{3/2}+\frac13b^{3/2}c^3 \\ &\ge\frac23a^{3/2}b^{1/2}c^{1/2}+\frac23a^{1/2}b^{3/2}c^{3/2}\\ &\ge \frac43abc. \end{align*}$$

Answer 3

I'd like to add a calculus approach to the above answers. If we define the function

$$ f(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} - abc $$

and look for the positions in $\mathbb{R}_{>0}^3$ where the gradient vanishes, we will find

$$ \overline{\triangledown } f = 0 \Rightarrow a=b=c=1 $$

The value of $f$ at $(1,1,1)$ is 0. However, the function $f$ happens to be "too symmetric" so that the Hessian matrix at $(1,1,1)$ has zero determinant, and thus the second derivative test gives no information about the nature of this critical point (the matrix has a zero eigenvalue).

Nevertheless, we can try a different "version" of the above function, putting $1/c$ in place of $c$.

$$ g(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{1}{6 c^6} - \dfrac{ab}{c} $$

Of course, this small transformation keeps (1,1,1) as the only critical point and leaves its nature unchanged. Checking the Hessian for this function at $(1,1,1)$ we get

$$H = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 2 & 1 \\ 1 & 1 & 7 \end{bmatrix}$$

which, using Sylvester's Criterion, is positive definite. Thus the point $(1,1,1)$ is a local minimum and since there is no other critical point in the area of interest, it is also a global minimum. That reveals that

$$ f(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} - abc >= 0 $$

for all $(a,b,c) \in \mathbb{R}_{>0}^3$

Answer 4

This can be solved using the weighted AM-GM inequality. For the case of three variables, this would say: $$x^{\alpha} y^{\beta} z^{\gamma} \le \alpha x + \beta y + \gamma z$$ given that $x,y,z > 0$; $\alpha, \beta, \gamma \ge 0$; and $\alpha + \beta + \gamma = 1$. Now, plug in $x = a^2$, $\alpha = \frac{1}{2}$, $y=b^3$, $\beta = \frac{1}{3}$, $z=c^6$, $\gamma = \frac{1}{6}$.

(Furthermore, if $\alpha, \beta, \gamma$ are all strictly positive, as in this application, then we have that equality holds if and only if $x = y = z$.)

Answer 5

I naturally want to generalize this. Here's a first cut.

$\begin{array}\\ \dfrac{a^u}{u}+\dfrac{b^v}{v} +\dfrac{c^{uv}}{uv} &=\dfrac{va^u+ub^v+c^{uv}}{uv}\\ &=\dfrac{va^u+ub^v+c^{uv}}{u+v+1}\dfrac{u+v+1}{uv}\\ &\ge (a^{uv}b^{uv}c^{uv})^{1/(u+v+1)}\dfrac{u+v+1}{uv}\\ &= (abc)^{uv/(u+v+1)}\dfrac{u+v+1}{uv}\\ &=\dfrac{(abc)^k}{k} \qquad k = \dfrac{uv}{u+v+1}\\ \end{array} $

with equality only if $a^u = b^v = c^{uv}$ or $a=c^v$ and $b=c^u$.

From now on I assume that $u \le v$.

If $k=1$, so that $uv = u+v+1$, or $(u-1)(v-1) = 2$, then $\dfrac{a^u}{u}+\dfrac{b^v}{v} +\dfrac{c^{uv}}{uv} \ge abc $.

If $uv = k(u+v+1)$, or $(u-k)(v-k) = k^2+k$, then $\dfrac{a^u}{u}+\dfrac{b^v}{v} +\dfrac{c^{uv}}{k} \ge \dfrac{(abc)^k}{k} $.

Since $k^2+k = k(k+1)$, two solutions to $(u-k)(v-k) = k^2+k$ are $u=2k, v=2k+1$ and $u=k+1, v=k^2+2k$.

To find all the integer solutions, write $k^2+k = rs$. Then $u=r+k, v=s+k$ are solutions. The ones above are for $r=1, s=k^2+k$ and $r=k, s=k+1$.

These are the same for $k=1$.

For $k=2$, $rs = 6$ so $(r, s) =(1, 6),(2, 3)$ and $(u, v) =(3, 8), (4, 5)$.

For $k=3$, $rs = 12$ so $(r, s) =(1, 12), (2, 6). (3, 4) $ and $(u, v) =(4, 15), (5, 9), (6, 7) $.

Answer 6

In general, if $$\sum_{i=1}^n \frac{1}{p_i} = 1$$ for positive $p_i$, then Jensen's inequality implies that for any concave function $\varphi$ defined on $\mathbb{R}^+$, we have $$\varphi\left(\sum_{i=1}^n \frac{x_i^{p_i}}{p_i}\right) \ge \sum_{i=1}^n \frac{1}{p_i} \varphi(x_i^{p_i}),$$ for positive $x_i$. In particular, if $\varphi(x) = \log x$, then we obtain Young's inequality: $$\sum_{i=1}^n \frac{x_i^{p_i}}{p_i} \ge \prod_{i=1}^n x_i.$$