Prove that $ \sum a^{b+c+d+e}>1$ $a,b,c,d,e>o$. Show that
$$ a^{b+c+d+e}+ b^{c+d+e+a}+c^{ d+e+a+b}+ d^{e+a+b+c}+e^{a+b+c+d}>1$$ 

Consider cases in which $a=b=c=d=e$. In such cases, the inequality (as I think you wanted to write it) simplifies to $5x^{4x} > 1$. To figure out if this is true, we need to consider the left side as a function of $x$, and find the vertex of this function. To do that, you need to take the derivative, $f'(x)$, and figure out where the derivative equals zero (and show that this vertex is a local minimum by demonstrating that $f'(x) < f(x\pm\delta)$ for small deviations $\delta$ from $x$), which makes this a calculus problem. Hint: for $y=x^x$, try taking the natural log of both sides first: $\ln(y) = \ln(x^x) = x \ln(x)$. Now, what's the derivative of a log?
I simply cheated and plugged it into a graphing calculator, and found that the vertex occurs at around $f(.36)=1.15$, so the inequality holds. For an exact answer, I leave the calculation of the derivative to you.
As per Calvin's comment, for cases where they are not all equal, any $x,y>1$ will mean that $x^y>1$, and any $x,y<1$ will mean $x,y<x^y<1$. The only way you can raise a base to a power and get a number smaller than either, is if $x<1<y$ or $y<1<x$, so in this inequality, you need five numbers < 1 but for which any four sum to > 1, thus making the result of each term smaller than each base and giving you the greatest chance for success.
You can show with a few examples that the higher the base, or the lesser the exponent, the larger the result, and vice-versa. Since every number will be a base and exponent at least once each, trying to minimize the base or maximize the exponent of any one term will backfire when the values you are manipulating swap places (for instance, $0.1^{10}=.0000000001$, but $10^{0.1}=1.2589$). 
The cases where all numbers are equal, eliminating any delta between terms, is therefore the ideal, and as I've shown above, there's no single value that would make the inequality false.
