Inequality with logarithms and radicals of order 4

Question

The statement of the problem : If $x,y \in (1, \infty)$ , prove that $\sqrt[4]{x} \bullet y^{\log_x y} \ge y$

My approach : If x = y , we have that $y^{\log_x y} = y$ , and with the fact that $x > 1$ $\implies$ $\sqrt[4]{x} > 1$ , and from here the inequality is obvious. If x < y $\implies$ $\log_x y >1$ , and proceeding as in the previous case, we prove the inequality. The problem came when I reached the case x > y . I tried to logarithmize base on x or y but it doesn't seem to lead anywhere.

Any and all proofs will be helpful. Thanks a lot!

Answer 1

In case of $x \gt y$, we have :

$y \lt x \implies \log_x y \lt 1 \implies \frac{1}{\log_y x} \gt 1 \implies y^{\frac{1}{\log_y x}} \gt y$.

Note here the inequality gets reversed in the second implication because $\log_x y$ is positive when $x,y \gt 1$. Now the solution follows from similar arguments as in $x \le y$.

Answer 2

This is the general solution where we don't necessarily have to order x and y.We are allowed to logarithmize to the base x without changing the meaning of the inequality because x,y>0. logarithmize in x base , x > 1 we get : $log_x (x^{1/4} * y^{log_x y}) \ge log_x y \iff$ $ 1/4 + (log_x y)^2 \ge log_x y \iff (2log_x y - 1)^2 \ge 0$ which is true for every $x,y \in (1, \infty).$