Algebra Equation solved by using AM-GM inequality

Exercise:

Solve in$\mathbb{R}$ , ${{16}^{x+{{y}^{2}}}}+{{16}^{y+{{x}^{2}}}}=1$

Solution: Apply the AM-GM inequality  $\frac{a+b}{2}\ge \sqrt{ab}\Leftrightarrow a+b\ge 2\sqrt{ab}$

$\Rightarrow {{\left( a+b \right)}^{2}}\ge 4ab$ So put $a={{16}^{x+{{y}^{2}}}}\,\,\And \,\,\,b={{16}^{{{x}^{2}}+y}}$

$\Rightarrow \left( {{16}^{x+{{y}^{2}}}}+{{16}^{{{x}^{2}}+y}} \right)\ge 4\left( {{16}^{x+{{y}^{2}}+{{x}^{2}}+y}} \right)$

$\Rightarrow 1\ge 4\left( {{16}^{x+{{y}^{2}}+{{x}^{2}}+y}} \right)\Rightarrow 1\ge 4\left( {{4}^{2x+2{{y}^{2}}+2{{x}^{2}}+2y}} \right)={{4}^{2x+2{{y}^{2}}+2{{x}^{2}}+2y+1}}$

So ${{4}^{0}}\ge {{4}^{2x+2{{y}^{2}}+2{{x}^{2}}+2y+1}}\Leftrightarrow 2x+2{{y}^{2}}+2{{x}^{2}}+2y+1\le 0$  Multiply by (2) to get

$\Rightarrow 4{{x}^{2}}+2{{y}^{2}}+4x+4y+2\le 0$ $\Rightarrow \left( {{\left( 2x \right)}^{2}}+2\left( 1 \right)\left( 2x \right)+1 \right)+\left( {{\left( 2y \right)}^{2}}+2\left( 1 \right)\left( 2y \right)+1 \right)\le 0$

$\Rightarrow {{\left( 2x+1 \right)}^{2}}+{{\left( 2y+1 \right)}^{2}}\le 0$ so ${{\left( 2x+1 \right)}^{2}}+{{\left( 2y+1 \right)}^{2}}=0$ thus $x=-\frac{1}{2}\,or\,y=-\frac{1}{2}$