Solving an exponential inequality $4^x + 25^x + 0.01^x \lt 2$ without using logarithms I need to solve this inequality without logarithms , 10th class


  $$4^x + 25^x + 0.01^x \lt 2$$


I got $$2^{2x} + 5^{2x} + 0.1^{2x} &lt;2$$ then $$5^{2x} + 0.1^{2x} &lt; 2(1-2^{2x-1}),$$ so after this I realized that the left side is greater than $0$, and so $1-2^{2x-1}$ should be greater than $0$ , and by logic I got that $x$ is less than $0.5$.  This is all I could do.
P.S thank u all
By AM-GM
$$4^x+25^x+\frac{1}{100^x}\geq3\sqrt[3]{4^x25^x\cdot\frac{1}{100^x}}=3&gt;2.$$
Can you end it now?
I think I realized how he wants you to do the problem but it is by no means obvious:

$4^x +25^x+10^{-2x} = 2^{2x}+5^{2x}+10^{-2x}= 2^{2x}+5^{2x}+5^{-2x}2^{-2x}$

Now let $u=2^{2x}$ and $v=5^{2x}$

Essentially you are solving the equation: 
$$ u+v +u^{-1}v^{-1} &lt; 2 \Rightarrow u+v&lt;2uv \Rightarrow u-2uv+v &lt;0 \Rightarrow (u-v)^2&lt;0$$

But this cannot be true for any real numbers u and v. Since I am assuming you are only searching for real solutions for x the same holds for $(2^{2x}-5^{2x})^2&lt;0$

And as such there are no real solutions.