<s>[INST] Prove that $a^a+b^b\ge a^b+b^a>1$ if $a,b> 0$. I succeeded in proving the second part of the inequality by showing that for every real number$x,y ∈ (0, 1)$, we have
$x^y≥ \frac x{x + y − xy}.$
By Bernoulli’s inequality we have
$x^{1−y}= (1 + x − 1)
^{1−y} 
   ≤ 1 + (x − 1)(1 − y) = x + y − xy,$
$\implies$ $x^y ≥ \frac x{x + y − xy}.$
If $a ≥ 1$ or $b ≥ 1$ then the given inequality clearly holds.
So let $0 <a,b< 1.$ By the previous inequality we have
$a^b + b^a≥ \frac a{a + b− ab}
+ \frac b{a + b− ab}= \frac {a+b}
{a + b− ab}
> \frac {a+b}{a+b}= 1.$
But, I found some difficulty on how to prove the first part. Thanks in advance
 [/INST] First, suppose $a>b>1$. Consider the function $f(x):=x^a-x^b$.
It suffices to show that $f$ is monotonically increasing on $[1,\infty[$.
But we can write $f(x)=x^b(x^{a-b}-1)$ which is the product of the two  strictly positive strictly increasing functions on $]1,\infty[$. So $f$ is monotonically increasing in this case.
Second, Suppose $a>1>b$. Since $a>1$ then $a^{x}$ is strictly increasing and we have $a^a>a^b$. And since $b<1$ then $b^x$ is strictly decreasing and we have that $b^a<b^b$.
Third, Suppose $1>a>b$. Write $f(x):=x^a-x^b=x^a(1-x^{b-a})$ then $f$ is an increasing function on $]0,1[$ (since it is again the product of two strictly positive increasing strictly functions).
The rest follows by symmetry
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