Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$

Question

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?

They all look like they want to be cubics... $\sqrt{a^3+b^3}\over \sqrt{a+b}$ for example... — abiessu, Oct 16 '13 at 04:58
@juantheron you can answer yourself if you don't feel confortable with my answer, but please don't leave this in the to-do queue for the rest of the eternity :) — chubakueno, Oct 29 '13 at 03:06

chubakueno · Answer 1 · 2013-10-16T05:52:41.020

5

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?

edited Oct 16 '13 at 05:52

answered Oct 16 '13 at 05:34

chubakueno

1 Answers1