Raising Itself

We will show that

$( \frac {1}{8} ) ^ { \frac {1}{8} } > ( \frac {1}{4} ) ^ { \frac {1}{4} }$

If we put both sides to the 8th power, we get:

$( \frac {1}{8} ) > ( \frac {1}{4} )^2 \Rightarrow \frac {1}{8} > \frac {1}{16}$

Since we found a counterexample, therefore the statement is not true in general.

$\text {Hence, our answer should be: } \boxed {FALSE}$

As another counterexample, we can use a calculator to shot at:

If x = 0.3 and y = 0.2, then x > y > 0 but $x^x < y^y$

$x^x = 0.3^{0.3} = 0.697 \text { (3 d. p.) and }$

$y^y = 0.2^{0.2} = 0.725 \text { (3 d. p.), therefore: }$

$x^x < y^y$

We can solve this without a calculator as well:

If we put both sides to the 10th power, we get:

$0.3^3 < 0.2^2 \iff 0.027 < 0.04$

If we want to see in general, when is our original statement true of false, we can use differentiation:

$x^x = e^{xlnx}$

After differentiating this, we get:

$e^{xlnx}(lnx +1)$

Since the first term (power of e) is always positive, therefore the first derivative is positive (and our original function is strictly monotonously increasing), when its second term is positive:

$lnx + 1 > 0$

$lnx > -1$

$x > \frac {1}{e}$

$\text {Which means, that } x^x > y^y \text { is always true, when: }$

$x > y ≥ \frac {1}{e}$

$\text {And it is always false (the opposite is true: } x^x < y^y \text { ), when: }$

$\frac {1}{e} ≥ x > y$

(In other cases, even equality is possible (e.g. x=0.5 , y = 0.25)

It is easy to prove that the inequality holds if

$x ≥ 1 > \frac {1}{e} ≥ y > 0$

If however:

$1 > x > \frac {1}{e} > y > 0$

then what we can say is (due to the facts, that $x^x \text { is continuous on the interval (0,1); }$

the limits (+ and - sides, respectively) at both of the two endpoints are equal to 1;

and the minimum point is between x and y):

$\forall x \exists y : x^x > y^y$

$\forall x \exists y : x^x < y^y$

$\forall x \exists y : x^x = y^y$

and vice versa:

$\forall y \exists x : x^x > y^y$

$\forall y \exists x : x^x < y^y$

$\forall y \exists x : x^x = y^y$ )

To prove that this statement is false, we just need a counter example. Observe that

$\left( \frac{1}{4} \right) ^ { \frac{1}{4} } > \left( \frac{1}{3} \right) ^ { \frac{1}{3} } \Leftrightarrow \left( \frac{1}{4} \right) ^ { 3 } > \left( \frac{1}{3} \right) ^ { 4 } \Leftrightarrow 3^4 > 4^3 \Leftrightarrow 81 > 64$

Hence, the statement is not true.

To determine the increasing / decreasing nature of $x^x$ , we can use calculus to show that it has a turning point at $x = \frac{1}{e}$ .

Draw the graph of x^x it is decreasing in interval (0,1/e) and in this interval given inequality does not hold whereas it is increasing in (1/e,infinity) and in this interval the given inequality holds. So this inequality is not true for all (x,y) satisfying x>y>0..