Minimize $A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$

Question

For $a,b>0$. Minimize $$A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$$

i think we let $2^x=a;2^y=b$

Hence $A=\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}$

We need pro $A\geq 2$(Wolfram Alpha) but $x,y$ is a very odd number and i can't find how to prove it $\geq2$

$$A=\frac{1+ab}{1+2a}+\frac{1+ab}{1+2b}$$ or $$A=\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}$$? — Khosrotash, Mar 15 '17 at 13:02
oh, if $A=\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}$ i know how to do it — Word Shallow, Mar 15 '17 at 13:22

score 3 · Answer 1 · answered Mar 15 '17 at 13:13

we have to prove that $$(1+ab)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\geq 2$$ and this is equivalent to $${\frac { \left( ab-1 \right) \left( a-b \right) ^{2}}{ \left( {a}^{2} +1 \right) \left( {b}^{2}+1 \right) }} \geq 0$$ this is right if $$ab\geq 1$$

Minimize $A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$

1 Answers1