Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$ Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$. 
By plotting graph I have seen that there are no roots for $x$. Can somebody prove it theoretically?  

Here's an answer, but I am sure there should be better ones. 
Assume $x \in \mathbb R$ to be a root of the given equation. Rearrange the equation as
$$
(x^{35}-100) (x^{17}+2+\cos^2 x) = - 20205.
$$
Let $A := x^{35}-100$ and $B := x^{17}+2+\cos^2 x$. Since $AB < 0$, exactly one of $A$ and $B$ is positive and the other is negative. 
If $B < 0$ and $A > 0$, then $x^{35} > 100 > 0$, which implies that $x > 0$. Then we can conclude that $B = x^{17}+2 + \cos^2 x > 0$, which contradicts the assumption about the sign of $B$. So this case is impossible.
We are left with the case $A < 0$ and $B > 0$. 


*

*From $A < 0$, we get $x^{35} < 100$. Therefore $x^{17} < 10$. Hence, $B = x^{17}+2+\cos^2 x \leq x^{17}+3 < 13$. 

*Similarly, from $B > 0$, we get $x^{17} > -2 - \cos^2 x \geq -3$. Therefore, $A = x^{35} - 100 > (-3)^{3}-100 = -127$. In other words, $|A| < 127$. 
Multiplying the upper bounds on $|A|$ and $B$, we get $$|A \cdot B| = |A| \cdot B < 127 \cdot 13 = 1651 ,$$
which contradicts the equation $AB = - 20205$ we started out with. 
