#### To determine

**To prove:** The relation cosh(sinhx)<sinh(cosh(x)).

**Solution:** We proved the inequality as shown below.

**Formulae used: **The hyperbolic property

#### Explanation

For any y≥0, notice

ey−1=∫0yexdx≥∫0y(1+x)dx≥∫0y(1+x1+x2)dx=y+1+y2−1

There is a little inequality:

1+y2−y=11+y2+y≥e−y

By using MVT, it could be found ξ∈(y,1+y2) such that:

sinh1+y2−sinh(y)=cosh(ξ)(1+y2−y)≥cosh(ξ)e−y≥e−y

**Since**e−y=cosh(y)−sinh(y), this will provide:

sinh1+y2≥cosh(y)

Substitute *y* by sinh(x) and notice 1+y2=cosh(x), this reduces to desired inequality and hence,

sinh(cosh(x))≥cosh(sin(x))