1Tried............ but din get NE thin close to wat U asked..........
x5-x3+x=k can be written as..........
x3[x2+(1/x2)+2-3]=k (x≠0) which is same as.............
x3[{x+(1/x)}2-3]=k
=> (k/x3)+3={x+(1/x)}2
When x>0
(k/x3)≥-1 or x3≥-k
When x<0
(k/x3)≥-5 or x3≤-k/5
ie., x3 ε [-k,-k/5]
Thats all wat I could figure out [2]
341k = x(x^4-x^2+1)
The second factor is always +ve. Hence the sign of k is the same as that of x.
If x≤0, then k≤0,and the inequality is obvious.
Let us look at the case x>0.
Then x^6+1 = (x^2+1)(x^4-x^2+1) \ge 2x(x^4-x^2+1) = 2k
From this we see that x^6 \ge 2k-1