# An Inequality with Exponents

### Problem

Dan Sitaru has kindly posted a problem from his book "Math Phenomenon" at the CutTheKnotMath facebook page. He also posted a solution (Solution 1).

If $a,b,c\in (0,1],\;$ then

$\displaystyle e^{\frac{4}{e}}\left(b\cdot a^{2\sqrt{a}}+c\cdot b^{2\sqrt{b}}+a\cdot c^{2\sqrt{c}}\right)\ge 3\sqrt[3]{abc}.$

When does the equality hold?

### Solution 1

Define $f(x):\;(0,1]\rightarrow\mathbb{R}\;$ with $f(x)=x^{2\sqrt{x}}.\;$ $f'(x)=\displaystyle x^{2\sqrt{x}-\frac{1}{2}}(2+\ln x).$

$\displaystyle\lim_{x\rightarrow 0^{+}}x^{2\sqrt{x}}=\lim_{x\rightarrow 0^{+}}e^{2\sqrt{x}\ln x}=e^{\displaystyle 2\lim_{x\rightarrow 0^{+}}\sqrt{x}\ln x}=e^0=1.$

$f'(x)$ has the only root that can be found from $2+\ln x=0,\;$ giving $x=e^{-2}.$ Thus $f(x)\;$ is monotone decreasing on $(0,e^{-2}]\;$ and monotone increasing on $(e^{-2},1]$. $f(1)=1.$ It follows that

$\displaystyle e^{-\frac{4}{e}}\le a^{2\sqrt{a}}\lt 1,\;$ $\displaystyle e^{-\frac{4}{e}}\le b^{2\sqrt{b}}\lt 1,\;$ $\displaystyle e^{-\frac{4}{e}}\le c^{2\sqrt{c}}\lt 1.\;$

And, subsequently,

$\displaystyle b\cdot e^{-\frac{4}{e}}\le ba^{2\sqrt{a}}\lt 1,\;$ $\displaystyle c\cdot e^{-\frac{4}{e}}\le cb^{2\sqrt{b}}\lt 1,\;$ $\displaystyle a\cdot e^{-\frac{4}{e}}\le ac^{2\sqrt{c}}\lt 1.\;$

Adding up and using the AM-GM inequality,

$\displaystyle b\cdot e^{-\frac{4}{e}}+c\cdot e^{-\frac{4}{e}}+a\cdot e^{-\frac{4}{e}}\ge (a+b+c)e^{-\frac{4}{e}}\ge 3\sqrt{abc}\cdot e^{-\frac{4}{e}}.$

In other words,

$\displaystyle e^{\frac{4}{e}}\left(b\cdot e^{-\frac{4}{e}}+c\cdot e^{-\frac{4}{e}}+a\cdot e^{-\frac{4}{e}}\right)\ge 3\sqrt{abc}.$

### Solution 2

This solution is much the same as the first one, with a few simplifications. First, replace $a=x^2,\;$ $b=y^2,\;$ $c=z^2\;$ to reduce the required inequality to

$\displaystyle e^{\frac{4}{e}}\left(y^2x^{4x}+z^2y^{4y}+x^2z^{4z}\right)\ge3\sqrt[3]{x^2y^2z^2}.$

With the AM-GM inequality we obtain

$\displaystyle\begin{align} \frac{1}{3}\left(y^2x^{4x}+z^2y^{4y}+x^2z^{4z}\right)&\ge \sqrt[3]{x^{4x}y^{4y}z^{4z}}\cdot\sqrt[3]{x^{2}y^{2}z^{2}}\\ &\ge \sqrt[3]{x^{2}y^{2}z^{2}}\sqrt[3]{\left[\left(\frac{1}{e}\right)^{\frac{1}{e}}\right]^{4\cdot 3}}\\ &= \sqrt[3]{x^{2}y^{2}z^{2}}\cdot\left(\frac{1}{e}\right)^{\frac{4}{e}}. \end{align}$

The latter inequality is the consequence of the properties of function $f(x)=x^x,\;$ defined for $x\gt 0.\;$ Its derivative $f'(x)=x^x(1+\ln x)\;$ vanishes only at $\displaystyle x=\frac{1}{e},\;$ where the function attains its minimum:

Indeed, the derivative $f'(x)=x^x(1+\ln x)\;$ is negative for $\displaystyle x\lt\frac{1}{e}\;$ and positive for $\displaystyle x\gt\frac{1}{e}.\;$ Thus the required inequality holds for $a,b,c\gt 0.$

The equality is attained when $\displaystyle x=y=z=\frac{1}{e},\;$ i.e., when $\displaystyle a=b=c=\frac{1}{e^2},\;$ in which case both sides of the inequality are equal to $3e^{-2}.$

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