Solving a System of Equations for Real Values of $a$ and $b$

[anemone]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

 

[mente oscura]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

Hello.

Spoiler

I do not know. At a glance:

$$a=1 \ and \ b=1$$

Regards.

 

[kaliprasad]

Hello.

Spoiler

I do not know. At a glance:

$$a=1 \ and \ b=1$$

Regards.

Spoiler

by interchanging a and b we get same set of equations so a= b is a solution set(that does not mean that other solution does not exist). taking a= b we get a=b = 1 is a solution set

 

[anemone]

Solve the following system for real values of $a$ and $b$:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

Suggested solution by J. Chui:

Spoiler

WLOG, let $a \ge b$ so that $\sqrt{a} \ge 1 \ge \sqrt{b}$.

Suppose that $\sqrt{a}=1+u$ and $\sqrt{b}=1-u$. Then $a+b=2+2u^2 \ge 2$ and $ab=(1-u^2)^2 \le 1$. Thus, by the AM-GM inequality, we have

$8=2^{a^2+b}+2^{a+b^2} \ge 2\sqrt{2^{a^2+b+a+b^2}}\ge 2 \sqrt{2^{(a+b)(a+b+1)-2ab}} \ge 2 \sqrt{2^{2\cdot3-2\cdot1}} \ge 2^3 \ge 8$

with equality iff $a=b$.

Since equality must hold throughtout, $a=b$ and thus the only solution to the system is $(a,b)=(1,1)$.