My solution:
For x<0, we have a positive left hand side value and a negative right hand side value. So x can never be a negative value.
For x>1, we have:
1+x2>2x,(1+x3)(1+x5)=1+x3+x5+x8>4x4 so (1+x2)(1+x3)(1+x5)>8x5, which really is 8x5>8x5, which leads to a contradiction.
For 0≤x≤1:
f(x)=(1+x2)(1+x3)(1+x5) has its first derivative of f′(x)>0 and so f is an increasing function and so does f(x)=8x5.
That means they can intersect at most once, and by inspection, it is not hard to see that x=1 is the only real solution to the system.
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