inequality

Tried............ 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 x³â‰¥-k

When x<0

(k/x³)≥-5 or x³â‰¤-k/5

ie., x³ ε [-k,-k/5]

Thats all wat I could figure out [2]

k = 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