Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\log_e x>0, \log_e y>0$, hence we can apply $AM>=GM$
$$\dfrac{log_e x+log_e y}{2}>=\sqrt{\log_e x\log_e y}$$
As both the sides are positive, so we can square both the sides without breaking the inequality.
$$(\log_e x)^2+(\log_e y)^2+2\log_e x\log_e y>=4\log_e x\log_e y$$
Using the given condition $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$
$$(\log_e x^2)+(\log_e y^2)>=2\log_e x\log_e y$$ $$(\log_e x)+(\log_e y)>=\log_e x\log_e y$$ $$\log_e xy>=\log_e x^{\log_e y}$$
As $e>1$, so we can safely write $xy>=x^{\log_e y}$
But actual answer is $e^4$, I am not able to think of any other way. Please help me in this.