Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$. Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$.

It's obvious that if  $x\gt 1$ then  $x+1\gt 2$ , also $x^5\gt0 \;as \;x\gt1\gt0$, thus $x^5+x+1\gt 2$.

Is it fine ?

I can't understand what do they mean by direct proof?
Since the derivative $f'(x)=5x^4+1$ is always positive, the function $f(x)=x^5+x+1$ is increasing over $\mathbb{R}$ and $x&gt;1$ implies $f(x)&gt;f(1)=3$.
The function is monotone increasing for positive values. This is because addition and multiplication are increasing in either argument. Moreover the value at 1 is 3. The theorem follows.