Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$ 
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\tag1\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1
$$

My Attempt:
Let $2^x = a$ and $3^x = b$ . Then $(1)$ becomes
$$
\sqrt{a^2-a\cdot b+b^2}+\sqrt{b^2-b+1}+\sqrt{a^2-a+1} = a+b+1
$$
How can I complete the solution from this point?

Hint:
$$a^2+b^2\ge 2ab$$
$$\implies 4a^2+4b^2-4ab\ge a^2+b^2+2ab$$
$$\implies \sqrt{a^2+b^2-ab}\ge \frac{a+b}{2}$$
With equality iff $a=b$
You see where this is going?
