Bleak Relationship #1

We note that $LHS > 0$ for $xy > 0$ . Then,

$\begin{aligned} 81^x+81^y + 3^{\frac{1}{x}} + 3^{\frac{1}{y}} & \ge 4 \sqrt[4]{81^x \dot{} 81^y \dot{} 3^{\frac{1}{x}} \dot{} 3^{\frac{1}{y}}} \\ & \ge 4 \sqrt[4]{3^{4x} \dot{} 3^{4y} \dot{} 3^{\frac{1}{x}} \dot{} 3^{\frac{1}{y}}} \end{aligned}$

Equality happens and $LHS$ is minimum when $\space 3^{4x} = 3^{4y} = 3^{\frac{1}{x}} = 3^{\frac{1}{y}}$

$\Rightarrow 4 x = \dfrac{1}{x} \quad \Rightarrow x^2 = \frac{1}{4} \quad \Rightarrow x = y = \frac{1}{2}$

$\begin{aligned} 81^x+81^y + 3^{\frac{1}{x}} + 3^{\frac{1}{y}} & \ge 4 \sqrt[4]{3^{2} \dot{} 3^{2} \dot{} 3^{2} \dot{} 3^{2}} \\ & \ge 36 \end{aligned}$

This implies that $81^x+81^y + 3^{\frac{1}{x}} + 3^{\frac{1}{y}} = 36$ if and only if $x = y = \frac{1}{2}$ and therefore, there is only $\boxed{1}$ solution.

3^(4x)+3^(1/x) >=2[3^(4x+1/x)]. By AM >=GM Further 4x+1/x >= 2[(4x)(1/x)]^1/2 4x+1/x >= 2 3^(4x)+3^(1/x) >=18

Equality holds for 4x=1/x x=1/2 , similarly y=1/2 Therefore for minimum sum of 36 equality holds and only ordered pair is (1/2, 1/2)