If $a, b$ and $c$ are real numbers such that $2^a + 4^b = 2^c$, $4^a + 2^b = 4^c$. Find the minimum value of $c$.
I've tried using arithmetic and geometric means and writing $2^a$ and $2^b$ in terms of each other and tried differentiating for the minimum value. I couldn't get any satisfactory results though. I've been trying this question from days but just can't figure out. Please explain what's supposed to be done here.
Set $$y=2^{a}, x=2^{b}, z=2^{c}.$$ Clearly $x,y,z>0,$ and $c$ is minimal when $z$ is so.
The given system rewrites $$y+x^2=z\tag1$$ $$y^2+x=z^2.\tag2$$ Square $(1)$ to obtain $$z^2=y^2+2yx^2+x^4,$$ equate with $z^2$ from $(2)$ and eliminate $y^2.$ We have $$2yx^2+x^4=x,$$ or equivalently, as $x\ne0,$ $$y=\frac{1-x^3}{2x}.$$ Putting this into $(1)$ we obtain $$z=\frac{x^3+1}{2x}.\tag3$$ Derivate $z(x)=z$ (for $x>0$ it is clearly differentiable) $$z'=\frac{2x^3-1}{2x}=0\quad\iff\quad {x=2^{-{1\over3}}}.$$ As $$z''=\frac{x^3+1}{x^3}>0,$$ the function $z$ is convex and $z(2^{-{1\over3}})$ is its minimum. Return to $(3)$ to obtain $$z_{min}=z(2^{-{1\over3}})=3\cdot 2^{-{5\over3}}$$ or $$2^{c}=2^{-{5\over3}+\log_2{3}}.$$ The minimum is $$c=-{5\over3}+\log_2{3}.$$